993 + Interview Questions in POLYNOMIALS IN GENERAL AWARENESS Page 12 InterviewSolution

551.	Factorise `x^(2)+(1)/(x^(2))-3`.
Answer» Given expression : `x^(2)+(1)/(x^(2))-3=x^(2)+(1)/(x^(2))-2-1=(x^(2)-2+(1)/(x^(2)))-1` `=(x-(1)/(x))^(2)-(1)^(2)=(x-(1)/(x)+a)(x-(1)/(x)-1)`

Discussion

552.	Factorise `16x^(4)-81y^(4)`.
Answer» Given expression : `16x^(4)-81y^(4)=(4x^(2))^(2)-(9y^(2))^(2)=(4x^(2)+9y^(2))(4x^(2)-9y^(2))` `=(4x^(2)+9y^(2))[(2x)^(2)-(3y)^(2)]` `=(4x^(2)+9y^(2))(2x+3y)(2x-3y)`

Discussion

553.	Factorise `x^(2)-6x-y^(2)+9`.
Answer» Given expression : `x^(2)-6x-y^(2)+9` Here, `x^(2)-6x+9` is perfect square. `therefore` By shifting the position of terms `x^(2)-6x-y^(2)+9=(x^(2)-6x+9)-y^(2)` `=(x-3)^(2)-(y)^(2) [because x^(2)-6x+9=(x)^(2)-2(x)(3)+(3)^(2)]` =(x-3+y)(x-3-y)=(x+y-3)(x-y-3)

Discussion

554.	Factorise `x^(2)-(y-z)^(2)`.
Answer» Given expression : `(x)^(2)-(y-z)^(2)=[x(y-z)][x-(y-z)] " " [because x^(2)-y^(2)=(x+y)(x-y)]` =(x+y-z)(x-y+z)

Discussion

555.	Factorise `a^(2)-9b^(2)`.
Answer» Given expression : `a^(2)=9b^(2)=(a)^(2)-(3b)^(2) " " [because x^(2)-y^(2)=(x+y)(x-y)]` =(a+3b)(a-3b)

Discussion

556.	Factorise a(x-y)-b(x-y).
Answer» Given expression : a(x-y)-b(x-y) Now, (x-y) is common in both terms. `therefore` First factor is (x-y). `therefore a(x-y)-b(x-y)=(x-y)(a-b)`

Discussion

557.	Write whether the following statements are true or false. Justify your answer.(i) A Binomial can have atmost two terms.(ii) Every polynomial is a Binomial.(iii) A binomial may have degree 5.(iv) Zero of a polynomial is always 0.(v) A polynomial cannot have more than one zero.(vi) The degree of the sum of two polynomials each of degree 5 is always 5.
Answer» (i) False, because a binomial has exactly two terms. (ii) False, because every polynomial is not a binomial . e.g., (a) 3x² + 4x + 5 [polynomial but hot a binomial] (b) 3x² + 5 [polynomial and also a binomial] (iii) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. So, it may have degree 5. (iv) False, because zero of a polynomial can be any real number e.g., p(x) = x – 2, then 2 is a zero of polynomial p(x). (v) False, because a polynomial can have any number of zeroes. It depends upon the degree of the polynomial e.g., p(x) = x² -2, as degree pf p(x) is 2 ,so it has two degree, so it has two zeroes i.e., √2 and —√2. (vi) False, because the sum of any two polynomials of same degree is not always same degree. e.g., Let f(x) = x⁴ + 2 and g(x) = -x⁴ + 4x³ + 2x ∴ Sum of two polynomials, f{x) + g(x) = x⁴ + 2 + (-x⁴ + 4x³ + 2x) = 4x³ + 2x + 2 which is not a polynomial of degree 4.

557.

Write whether the following statements are true or false. Justify your answer.(i) A Binomial can have atmost two terms.(ii) Every polynomial is a Binomial.(iii) A binomial may have degree 5.(iv) Zero of a polynomial is always 0.(v) A polynomial cannot have more than one zero.(vi) The degree of the sum of two polynomials each of degree 5 is always 5.

Answer»

(i) False, because a binomial has exactly two terms.

(ii) False, because every polynomial is not a binomial .
e.g., (a) 3x² + 4x + 5 [polynomial but hot a binomial]

(b) 3x² + 5 [polynomial and also a binomial]

(iii) True, because a binomial is a polynomial whose degree is a whole number greater than equal to one. So, it may have degree 5.

(iv) False, because zero of a polynomial can be any real number e.g., p(x) = x – 2, then 2 is a zero of polynomial p(x).

(v) False, because a polynomial can have any number of zeroes. It depends upon the degree of the polynomial
e.g., p(x) = x² -2, as degree pf p(x) is 2 ,so it has two degree, so it has two zeroes i.e., √2 and —√2.

(vi) False, because the sum of any two polynomials of same degree is not always same degree.
e.g., Let f(x) = x⁴ + 2 and g(x) = -x⁴ + 4x³ + 2x

∴ Sum of two polynomials,

f{x) + g(x) = x⁴ + 2 + (-x⁴ + 4x³ + 2x)

= 4x³ + 2x + 2

which is not a polynomial of degree 4.

Discussion

558.	Factorise `p^(2)x^(2)+c^(2)x^(2)-ac^(2)-ap^(2)`.
Answer» Given expression : `p^(2)x^(2)+c^(2)x^(2)-ac^(2)-ap^(2)` Here, `x^(2)` is common in first two terms and (-a) is common in last two terms. `therefore p^(2)x^(2)+c^(2)x^(2)-ac^(2)-ap^(2)=x^(2)(p^(2)+c^(2))-a(c^(2)+p^(2))` `=x^(2)(p^(2)+c^(2))-a(p^(2)+c^(2))=(p^(2)+c^(2))(x^(2)-a)`

Discussion

559.	Factorise `x^(2)-(a+(1)/(a))x+1`.
Answer» Given expression : `x^(2)-(a+(1)/(a))x+1` `x^(2)-ax-(x)/(a)+1=x^(2)-ax-(x)/(a)+(a)/(a)` `=x(x-a)-(1)/(a)(x-a)=(x-a)(x-(1)/(a))`

Discussion

560.	If a + b + c = 0, then a3 + b3 + c3 is equal to (A) 0 (B) abc (C) 3abc (D) 2abc
Answer» (c) 3abc Now, a³+b³ + c³= (a+ b + c) (a² + b² + c² – ab – be – ca) + 3abc [using identity, a³+b³ + c³ – 3 abc = (a + b + c)(a² + b² + c² – ab – be -ca)] = 0 + 3abc [∴ a + b + c = 0, given] a³+b³ + c³ = 3abc

560.

If a + b + c = 0, then a3 + b3 + c3 is equal to (A) 0 (B) abc (C) 3abc (D) 2abc

Answer»

(c) 3abc

Now, a³+b³ + c³= (a+ b + c) (a² + b² + c² – ab – be – ca) + 3abc
[using identity, a³+b³ + c³ – 3 abc = (a + b + c)(a² + b² + c² – ab – be -ca)]

= 0 + 3abc [∴ a + b + c = 0, given]

a³+b³ + c³ = 3abc

Discussion

561.	Write zeros of the polynomial `p(x)=4sqrt(3)x^2+5x-2sqrt(3)`
Answer» Correct Answer - ` x = (-2)/sqrt3 or x = sqrt3/4 ` `f(x) = 4 sqrt3 x^(2) + 5x -2 sqrt3` `=4sqrt3 x^(2) + 8x-3x - 2sqrt3 = 4x(sqrt3x + 2) -sqrt3(sqrt3(sqrt3 x + 2)` ` = (sqrt3 x+2)(4x-sqrt3).` So, its zeros are `(-2)/sqrt3, sqrt3/4.`

Discussion

562.	Factorise ax+ay+az+bx+by+bz.
Answer» ax+ay+az+bx+by+bz=(ax+ay+az)+(bx+by+bz) =a(x+y+z)+b(x+y+z)=(x+y+z)(a+b)

Discussion

563.	The coefficient of x in the expansion of (x + 3)3 is (A) 1 (B) 9 (C) 18 (D) 27
Answer» (d) 27 Now, (x + 3)³ = x³ + 3³ + 3x (3)(x + 3) [using identity, (a + b)³ = a³ + b³ + 3ab (a + b)] = x³ + 27 + 9x (x + 3) = x³ +27 + 9x²+27x Hence, the coefficient of x in (x + 3)³ is 27.

563.

The coefficient of x in the expansion of (x + 3)3 is (A) 1 (B) 9 (C) 18 (D) 27

Answer»

(d) 27

Now, (x + 3)³ = x³ + 3³ + 3x (3)(x + 3)
[using identity, (a + b)³ = a³ + b³ + 3ab (a + b)]

= x³ + 27 + 9x (x + 3)

= x³ +27 + 9x²+27x

Hence, the coefficient of x in (x + 3)³ is 27.

Discussion

564.	If ` p(x)=x^(2)-2sqrt(2x)+1` , then p `(2sqrt(2))` is equal to
Answer» Correct Answer - B Given , ` " "p(x)=x^(2)=2sqrt(2)x+1 " "` On putting `x=2sqrt(2)`in Eq ,(i) We get `p(2sqrt(2))=(2sqrt(2))^(2)-(2sqrt(2)) (2sqrt(2))+1` `=8-8+1=1`

Discussion

565.	Factorise the following expressions : `(i) ax-ay+bx-by " " (ii) x^(2)-x-ax+a " " (iii) x^(4)+x^(3)+x^(2)+x` `(iv) 16(a+b)^(2)-4a-4b " " (v) x^(2)+(1)/(x^(2))+2-3x-(3)/(x) " " (vi) x^(2)-((a)/(b)+(b)/(a))x+1` `(vii) x^(2)+(a-(1)/(a))x-1 " " (viii)ab(x^(2)+y^(2)+xy(a^(2)+b^(2)) " "(ix) (ax+by)^(2)+(bx-ay)^(2)`
Answer» Correct Answer - `(i) (x-y)(a+b) " " (ii) (x-1)(x-a) " " (iii) x(x+1)(x^(2)+1)` (iv)` 4(a+b)(4a+4b-1) " " (v) (x+(1)/(x))(x+(1)/(1)/(x)-3) " "(vi) (x-(a)/(b))(x-(b)/(a))` `(vii) (a-a)(x-(1)/(a)) " " (viii) (bx+ay)(ax+by) " " (viii) (x^(2)+y^(2))(a^(2)+b^(2))`

Discussion

566.	the factorisation of `4x^(2)+8x+3` isA. `(x+1)(x+3)`B. `(2x+1)(2x+3)`C. `(2x+2)(2x+5)`D. `(2x-1)(2x-3)`
Answer» Correct Answer - B Now , `4x^(2)+8x+3=4x^(2)+6x+2x+3 " " ["by splitting middle term"]` `=2x(2x+3)+1(2x+3)` `=(2x+3)(2x+1)`

Discussion

567.	Which of the following is a factor of (x + y)3 – (x3 + y3 )? (A) x2 + y2 + 2xy (B) x2 + y2 – xy (C) xy2(D) 3xy
Answer» (d) 3xy Now, (x+ y)³ – (x³ + y³) = (x + y) – (x + y)(x²– xy + y²) [using identity, a³ + b³ = (a + b)(a² -ab+ b²)] = (x+ y)[(x+ y)² -(x² -xy+ y²)] = (x+ y)(x²+ y²+ 2xy- x²+ xy- y²) [using identity, (a + b)² = a² + b² + 2 ab)] = (x + y) (3xy) Hence, one of the factor of given polynomial is 3xy.

567.

Which of the following is a factor of (x + y)3 – (x3 + y3 )? (A) x2 + y2 + 2xy (B) x2 + y2 – xy (C) xy2(D) 3xy

Answer»

(d) 3xy

Now, (x+ y)³ – (x³ + y³)

= (x + y) – (x + y)(x²– xy + y²)
[using identity, a³ + b³ = (a + b)(a² -ab+ b²)] = (x+ y)[(x+ y)² -(x² -xy+ y²)]

= (x+ y)(x²+ y²+ 2xy- x²+ xy- y²)
[using identity, (a + b)² = a² + b² + 2 ab)]

= (x + y) (3xy)

Hence, one of the factor of given polynomial is 3xy.

Discussion

568.	the value of `249^(2)-248^(2)` isA. `1^(2)`B. 477C. 487D. 497
Answer» Correct Answer - D Now `249^(2)-248^(2)=(249+248)(249-248) " " ["using identity ," a^(2)-b^(2)=(a-b)(a+b)]` `497xx1=497`

Discussion

569.	The factorisation of 4x2 + 8x + 3 is (A) (x + 1) (x + 3) (B) (2x + 1) (2x + 3) (C) (2x + 2) (2x + 5) (D) (2x –1) (2x –3)
Answer» (b) (2x+1) (2x + 3) Now, 4x² + 8x + 3 = 4x² + 6x + 2x + 3 [by splitting middle term] = 2x(2x+ 3) + 1 (2x+ 3) = (2x + 3) (2x + 1)

569.

The factorisation of 4x2 + 8x + 3 is (A) (x + 1) (x + 3) (B) (2x + 1) (2x + 3) (C) (2x + 2) (2x + 5) (D) (2x –1) (2x –3)

Answer»

(b) (2x+1) (2x + 3)

Now, 4x² + 8x + 3

= 4x² + 6x + 2x + 3 [by splitting middle term]

= 2x(2x+ 3) + 1 (2x+ 3)

= (2x + 3) (2x + 1)

Discussion

570.	If p(x) = x + 3, then p(x) + p(–x) is equal to(A) 3(B) 2x(C) 0(D) 6
Answer» (D) 6 Explanation: p(x) = x + 3 p(– x) = – x + 3 Therefore, p(x) + p(–x) = (x + 3) + (– x + 3) = x + 3 – x + 3 = 6 Hence, option D is the correct answer

570.

If p(x) = x + 3, then p(x) + p(–x) is equal to(A) 3(B) 2x(C) 0(D) 6

Answer»

(D) 6

Explanation:

p(x) = x + 3

p(– x) = – x + 3

Therefore,

p(x) + p(–x) = (x + 3) + (– x + 3)

= x + 3 – x + 3

= 6

Hence, option D is the correct answer

Discussion

571.	The value of 2492 – 2482 is (A) 12 (B) 477 (C) 487 (D) 497
Answer» (d) 497 Now, 249² – 248² = (249 + 248) (249 – 248) [using identity, a² – b² = (a – b)(a + b)] = 497 x 1 = 497.

571.

The value of 2492 – 2482 is (A) 12 (B) 477 (C) 487 (D) 497

Answer»

(d) 497

Now, 249² – 248²

= (249 + 248) (249 – 248) [using identity, a² – b² = (a – b)(a + b)]

= 497 x 1

= 497.

Discussion

572.	Zero of the zero polynomial is(A) 0(B) 1(C) Any real number(D) Not defined
Answer» (C) Any real number Explanation: Zero polynomial is a constant polynomial whose coefficients are all equal to 0. Zero of a polynomial is the value of the variable that makes the polynomial equal to zero. Therefore, zero of the zero polynomial is any real number. Hence, option C is the correct answer

572.

Zero of the zero polynomial is(A) 0(B) 1(C) Any real number(D) Not defined

Answer»

(C) Any real number

Explanation:

Zero polynomial is a constant polynomial whose coefficients are all equal to 0.

Zero of a polynomial is the value of the variable that makes the polynomial equal to zero.

Therefore, zero of the zero polynomial is any real number.

Hence, option C is the correct answer

Discussion

573.	One of the factors of (25x2 – 1) + (1 + 5x)2 is (A) 5 + x (B) 5 – x (C) 5x – 1 (D) 10x
Answer» (d) 10x Now, (25x² -1) + (1 + 5x)² = 25x² -1 + 1 + 25x² + 10x [using identity, (a + b)² = a² + b² + 2ab] = 50x² + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

573.

One of the factors of (25x2 – 1) + (1 + 5x)2 is (A) 5 + x (B) 5 – x (C) 5x – 1 (D) 10x

Answer»

(d) 10x

Now, (25x² -1) + (1 + 5x)²

= 25x² -1 + 1 + 25x² + 10x [using identity, (a + b)² = a² + b² + 2ab]

= 50x² + 10x

= 10x (5x+ 1)

Hence, one of the factor of given polynomial is 10x.

Discussion

574.	If `x+1` is a factor of the polynomial `2x^(2)+kx,` then the value of `k` isA. `-3`B. 4C. 2D. `-2`
Answer» Correct Answer - C Let ` " "p(x)=2x^(2)+kx` Since ,(x+1) is a factor of p(x) then p(-1) =0 `therefore " " 2(-1)^(2)+k(-1)=0` `implies " " 2 -l=0implies k=2` Hence , the value of k is 2.

Discussion

575.	x + 1 is a factor of the polynomial(a) x3 + x2 – x +1(b) x3 + x2 + x+1(c) x4 + x3 + x2 +1(d) x4 + 3x3 + 3x2 + x +1
Answer» (b) x³ + x² + x+1 Let assume (x + 1) is a factor of x³ + x² + x+1. So, x = -1 is zero of x³ + x² + x+1 (-1)³ + (-1)² + (-1) + 1 = 0 => -1+1-1 + 1 = 0 => 0 = 0 Hence, our assumption is true.

575.

x + 1 is a factor of the polynomial(a) x3 + x2 – x +1(b) x3 + x2 + x+1(c) x4 + x3 + x2 +1(d) x4 + 3x3 + 3x2 + x +1

Answer»

(b) x³ + x² + x+1

Let assume (x + 1) is a factor of x³ + x² + x+1.

So, x = -1 is zero of x³ + x² + x+1

(-1)³ + (-1)² + (-1) + 1 = 0

=> -1+1-1 + 1 = 0

=> 0 = 0

Hence, our assumption is true.

Discussion

576.	Zero of the polynomial p(x) = 2x + 5 is(A) – 2/5(B) – 5/2(C) 2/5(D) 5/2
Answer» (B) – 5/2 Explanation: Zero of the polynomial ⇒ p(x) = 0 p(x) = 0 2x + 5 = 0 2x = – 5 x = – 5/2 Hence, option B is the correct answer

576.

Zero of the polynomial p(x) = 2x + 5 is(A) – 2/5(B) – 5/2(C) 2/5(D) 5/2

Answer»

(B) – 5/2

Explanation:

Zero of the polynomial ⇒ p(x) = 0

p(x) = 0

2x + 5 = 0

2x = – 5

x = – 5/2

Hence, option B is the correct answer

Discussion

577.	If `x^(51) +51` is divided by `x+1` , then the remainder is
Answer» Correct Answer - D Let `p(x) =x^(51)+51` when we divide `p(x) " by " x+1 ` ,we get the remainder p(-1) On putting `x=-1` in Eq,(i) we get `p(-1) =(-1)^(51) +51` `=-1+51=50` Hence , the remainder is 50.

Discussion

578.	One of the zeroes of the polynomial 2x2 + 7x –4 is(A) 2(B) ½(C) – ½(D) –2
Answer» (B) ½ Explanation: Zero of the polynomial ⇒ p(x) = 0 p(x) = 0 2x² + 7x – 4 = 0 2x² – 1x + 8x – 4 = 0 x(2x – 1) + 4(2x – 1) = 0 (x + 4)(2x – 1) = 0 Consider, x + 4 x + 4 = 0 x = – 4 Consider, 2x – 1 2x – 1 = 0 2x = 1 x = ½ Hence, option B is the correct answer

578.

One of the zeroes of the polynomial 2x2 + 7x –4 is(A) 2(B) ½(C) – ½(D) –2

Answer»

(B) ½

Explanation:

Zero of the polynomial ⇒ p(x) = 0

p(x) = 0

2x² + 7x – 4 = 0

2x² – 1x + 8x – 4 = 0

x(2x – 1) + 4(2x – 1) = 0

(x + 4)(2x – 1) = 0

Consider, x + 4

x + 4 = 0

x = – 4

Consider, 2x – 1

2x – 1 = 0

2x = 1

x = ½

Hence, option B is the correct answer

Discussion

579.	If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is (A) –3 (B) 4 (C) 2 (D) –2
Answer» (c) 2 Let p(x) = 2x² + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 => 2-k = 0 => k= 2 Hence, the value of k is 2.

579.

If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is (A) –3 (B) 4 (C) 2 (D) –2

Answer»

(c) 2

Let p(x) = 2x² + kx

Since, (x + 1) is a factor of p(x), then

p(-1)=0

2(-1)2 + k(-1) = 0

=> 2-k = 0 => k= 2

Hence, the value of k is 2.

Discussion

580.	If x51 + 51 is divided by x + 1, then the remainder is(a) 0(b)(c) 49(d) 50
Answer» (d) 50 Let p(x) = x⁵¹ + 51 . …(i) When we divide p(x) by x+1, we get the remainder p(-1) On putting x= -1 in Eq. (i), we get p(-1) = (-1)⁵¹ + 51 = -1 + 51 = 50 Hence, the remainder is 50.

580.

If x51 + 51 is divided by x + 1, then the remainder is(a) 0(b)(c) 49(d) 50

Answer»

(d) 50

Let p(x) = x⁵¹ + 51 . …(i)

When we divide p(x) by x+1, we get the remainder p(-1)

On putting x= -1 in Eq. (i), we get

p(-1) = (-1)⁵¹ + 51

= -1 + 51

= 50

Hence, the remainder is 50.

Discussion

581.	Evaluate: (107)2
Answer» = (100 + 7)² = (100)² + (7)² + 2 × 100 × 7 = 10000 + 49 + 1400 = 11449

581.

Evaluate: (107)2

Answer»

= (100 + 7)²

= (100)² + (7)² + 2 × 100 × 7

= 10000 + 49 + 1400

= 11449

Discussion

582.	Evaluate:(94)2
Answer» = (100 - 6)² = (100)2 + (6)² - 2 × 100 × 6 = 10000 + 36 - 1200 = 8836

582.

Evaluate:(94)2

Answer»

= (100 - 6)²

= (100)2 + (6)² - 2 × 100 × 6

= 10000 + 36 - 1200

= 8836

Discussion

583.	Evaluate the following using suitable identities:(99)3
Answer» It is known that, (a + b)³ = a³ + b³ + 3ab(a + b) and (a- b)³= a³ - b³ -3ab(a - b) (99)³ = (100 − 1)³ = (100)³ − (1)³ − 3(100) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299

583.

Evaluate the following using suitable identities:(99)3

Answer»

It is known that,

(a + b)³ = a³ + b³ + 3ab(a + b)

and (a- b)³= a³ - b³ -3ab(a - b)

(99)³ = (100 − 1)³

= (100)³ − (1)³ − 3(100) (1) (100 − 1)

= 1000000 − 1 − 300(99)

= 1000000 − 1 − 29700

= 970299

Discussion

584.	Evaluate:(0.99)2
Answer» = (1 - 0.01)² = (1)² + (0.01)² - 2 × 1 × 0.01 = + 0.0001 - 0.02 = 0.9801

584.

Evaluate:(0.99)2

Answer»

= (1 - 0.01)²

= (1)² + (0.01)² - 2 × 1 × 0.01

= + 0.0001 - 0.02

= 0.9801

Discussion

585.	Simplify : (3x + 4)3 - (3x - 4)3 .
Answer» We have........... (3x + 4)³ - (3x - 4)³ = [(3x)³ + (4)³ + 3 × 3x × 4 × (3x + 4)] - [(3x)³ - (4)³ - 3 × 3x × 4 × (3x - 4)] = [27³ + 64 + 36x (3x + 4)] - [27³ - 64 - 36x (3x - 4)] = [27x³ + 64 + 108x² + 144x] - [27x³ - 64 - 108x² + 144x] = 27x³ + 64 + 108x² + 144x - 27x³ + 64 + 108x² - 144x = 128 + 216x² ∵ (3x + 4)³ - (3x - 4)³ = 128 + 216x²

585.

Simplify : (3x + 4)3 - (3x - 4)3 .

Answer»

We have...........

(3x + 4)³ - (3x - 4)³ = [(3x)³ + (4)³ + 3 × 3x × 4 × (3x + 4)] - [(3x)³ - (4)³ - 3 × 3x × 4 × (3x - 4)]

= [27³ + 64 + 36x (3x + 4)] - [27³ - 64 - 36x (3x - 4)]

= [27x³ + 64 + 108x² + 144x] - [27x³ - 64 - 108x² + 144x]

= 27x³ + 64 + 108x² + 144x - 27x³ + 64 + 108x² - 144x

= 128 + 216x²

∵ (3x + 4)³ - (3x - 4)³ = 128 + 216x²

Discussion

586.	Evaluate: (1005)3
Answer» = (1000 + 5)³ = (1000)³ + (5)³ + 3 × 1000 × 5 × (1000 + 5) = 1000000000 + 125 + 15000 + (1000 + 5) = 1000000000 + 125 + 15000000 + 75000 = 1015075125

586.

Evaluate: (1005)3

Answer»

= (1000 + 5)³

= (1000)³ + (5)³ + 3 × 1000 × 5 × (1000 + 5)

= 1000000000 + 125 + 15000 + (1000 + 5)

= 1000000000 + 125 + 15000000 + 75000

= 1015075125

Discussion

587.	Evaluate:(997)3
Answer» = (1000 - 3)³ = (1000)³ - (3)³ - 3 × 1000 × 3 × (1000 - 3) = 1000000000 - 27 - 9000 × (1000 - 3) = 1000000000 - 27 - 900000 + 27000 = 991026973

587.

Evaluate:(997)3

Answer»

= (1000 - 3)³

= (1000)³ - (3)³ - 3 × 1000 × 3 × (1000 - 3)

= 1000000000 - 27 - 9000 × (1000 - 3)

= 1000000000 - 27 - 900000 + 27000

= 991026973

Discussion

588.	Find the following products of the following expression: (4x + 3y) (16x2 - 12xy + 9y2 )
Answer» (4x + 3y) (16x² - 12xy + 9y² ) = (4x + 3y) [(4x)² - (4x) × (3y) + (3y)² ] = (x + b) (x² - ab + b² ) [Where a = 4x, b = 3y] = a³ + b³ = (4x)³ + (3y)³ = 64x³ + 27y³

588.

Find the following products of the following expression: (4x + 3y) (16x2 - 12xy + 9y2 )

Answer»

(4x + 3y) (16x² - 12xy + 9y² )

= (4x + 3y) [(4x)² - (4x) × (3y) + (3y)² ]

= (x + b) (x² - ab + b² ) [Where a = 4x, b = 3y]

= a³ + b³

= (4x)³ + (3y)³

= 64x³ + 27y³

Discussion

589.	Find the following products of the following expression:(5x - 2y) (25x2 + 10xy + 4y2 )
Answer» (5x - 2y) (25x² + 10xy + 4y² ) = (5x - 2y) [(5x² + (5x) × (2y) + (2y)² ] = (a - b) (a² + ab + b² ) [Where a = 5x, b = 2y] = a³ - b³ = (5x)³ - (2y)³ = 125x³ - 8y³

589.

Find the following products of the following expression:(5x - 2y) (25x2 + 10xy + 4y2 )

Answer»

(5x - 2y) (25x² + 10xy + 4y² )

= (5x - 2y) [(5x² + (5x) × (2y) + (2y)² ]

= (a - b) (a² + ab + b² ) [Where a = 5x, b = 2y]

= a³ - b³

= (5x)³ - (2y)³

= 125x³ - 8y³

Discussion

590.	Prove that : (x - y)3 + (y - z)3 + (z - x)3 = 3(x - y) (y - z) (z - x).
Answer» Let (x - y) = a, (y - z) = b and (z - x) = c. Then, a + b + c = (x - y) + (y - z) + (z - x) = 0 ∴a ³ + b³ + c³ = 3abc Or (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y) (y - z) (z - x)

590.

Prove that : (x - y)3 + (y - z)3 + (z - x)3 = 3(x - y) (y - z) (z - x).

Answer»

Let (x - y) = a, (y - z) = b and (z - x) = c.

Then, a + b + c = (x - y) + (y - z) + (z - x) = 0

∴a ³ + b³ + c³ = 3abc

Or (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y) (y - z) (z - x)

Discussion

591.	Find the value of (28)3 - (78)3 + (50)3
Answer» Let a = 28, b = - 78, c = 50 Then, a + b + c = 28 - 78 + 50 = 0 ∴ a³ + b³ + c³ = 3abc. So, (28)³ + (-78)³ + (50)³ = 3 × 28 × (-78) × 50

591.

Find the value of (28)3 - (78)3 + (50)3

Answer»

Let a = 28, b = - 78, c = 50

Then, a + b + c = 28 - 78 + 50 = 0

∴ a³ + b³ + c³ = 3abc.

So, (28)³ + (-78)³ + (50)³ = 3 × 28 × (-78) × 50

Discussion

592.	If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 - 3abc.
Answer» We have a + b + c = 9 ...(i) ⇒ (a + b + c)² = 81 [On squaring both sides of (i)] ⇒ a² + b² + c² + 2(ab + bc + ac) = 81 ⇒ a² + b² + c² + 2 × 26 = 81 [∵ab + bc + ac = 26] ⇒ a² + b² + c² = (81 - 52) ⇒ a² + b² + 2 = 29. Now, we have a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac) = (a + b + c) [(a² + b² + c² ) - (ab + bc + ac)] = 9 × [(29 - 26)] = (9 × 3) = 27

592.

If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 - 3abc.

Answer»

We have a + b + c = 9 ...(i)

⇒ (a + b + c)² = 81 [On squaring both sides of (i)]

⇒ a² + b² + c² + 2(ab + bc + ac) = 81

⇒ a² + b² + c² + 2 × 26 = 81 [∵ab + bc + ac = 26]

⇒ a² + b² + c² = (81 - 52)

⇒ a² + b² + 2 = 29.

Now, we have

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)

= (a + b + c) [(a² + b² + c² ) - (ab + bc + ac)]

= 9 × [(29 - 26)]

= (9 × 3)

= 27

Discussion

593.	Find a quadratic polynomial with zeroes -2 and 1/3.
Answer» Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β. Here α = – 2 and β = 1/3 Sum of the zeroes = α + β = -2 + 1/3 = -5/3 Product of the zeroes = αβ = 1/3 × (-2) = -2/3 ∴ ax² + bx + c is [x² – (α + β)x + αβ] = [ x²- (-5/3) x + (-2/3)] the quadratic polynomial will be 3x² + 5x – 2.

593.

Find a quadratic polynomial with zeroes -2 and 1/3.

Answer»

Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.

Here α = – 2 and β = 1/3

Sum of the zeroes = α + β

= -2 + 1/3 = -5/3

Product of the zeroes = αβ

= 1/3 × (-2) = -2/3

∴ ax² + bx + c is [x² – (α + β)x + αβ]

= [ x²- (-5/3) x + (-2/3)]

the quadratic polynomial will be 3x² + 5x – 2.

Discussion

594.	Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 + 2x + 1
Answer» Given polynomial p(x) = x² + 2x + 1 We have x² + 2x + 1 = x² + x + x + 1 = x(x + 1) + l(x + 1) = (x + 1) (x + 1) So, the value of x² + 2x + 1 is zero when x + 1 = 0 (or) x + 1 = 0, i.e., when x = – 1 or – 1 So, the zeroes of x² + 2x + 1 are – 1 and – 1. ∴ Sum of the zeroes = (-1) + (-1) = -2 \(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-2}{1}={-2}\) And product of the zeroes = (-1) × (-1) = 1 \(=\frac{Constant}{Coefficient\,of\,x^2}=\frac{1}{1}=1\)

594.

Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 + 2x + 1

Answer»

Given polynomial p(x) = x² + 2x + 1

We have x² + 2x + 1 = x² + x + x + 1

= x(x + 1) + l(x + 1)

= (x + 1) (x + 1)

So, the value of x² + 2x + 1 is zero

when x + 1 = 0 (or) x + 1 = 0, i.e.,

when x = – 1 or – 1

So, the zeroes of x² + 2x + 1 are – 1 and – 1.

∴ Sum of the zeroes = (-1) + (-1) = -2

\(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-2}{1}={-2}\)

And product of the zeroes = (-1) × (-1) = 1

\(=\frac{Constant}{Coefficient\,of\,x^2}=\frac{1}{1}=1\)

Discussion

595.	Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 – 4
Answer» Given polynomial p(x) = x² – 4 We have, x² – 4 = (x – 2) (x + 2) So, the value of x² – 4 is zero when x – 2 = 0 or x + 2 = 0 i.e., x = 2 or x = – 2 So the zeroes of x² – 4 are 2 and – 2 ∴ Sum of the zeroes = 2 + (- 2) = 0 \(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}= \frac{-0}{1}=1\) And product of the zeroes = 2 × (-2) = -4 \(=\frac{Consatnt\,term}{Coefficient\,of\,x^2}=\frac{-4}{1}=-4\)

595.

Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 – 4

Answer»

Given polynomial p(x) = x² – 4

We have, x² – 4 = (x – 2) (x + 2)

So, the value of x² – 4 is zero

when x – 2 = 0 or x + 2 = 0

i.e., x = 2 or x = – 2

So the zeroes of x² – 4 are 2 and – 2

∴ Sum of the zeroes = 2 + (- 2) = 0

\(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}= \frac{-0}{1}=1\)

And product of the zeroes = 2 × (-2) = -4

\(=\frac{Consatnt\,term}{Coefficient\,of\,x^2}=\frac{-4}{1}=-4\)

Discussion

596.	Find the zeroes of the polynomial `x^2-3`and verify the relationship between the zeroes and the coefficients.
Answer» `P(x)=x^2-3` `x^2-3=0` `(x)^2-(sqrt3)^2` `(x-sqrt3)(x+sqrt3)=0` `x=sqrt3,-sqrt3` `alpha=sqrt3,beta=-sqrt3` `alpha+beta=sqrt3-sqrt3=0` `alphabeta=sqrt3(-sqrt3)=-3` `ax^2+bx+c=0` `alpha+beta=-b/a=-(0)/1=0` `alphabeta=c/a=(-3)/1=-3`

Discussion

597.	Find a quadratic polynomial, the sum and product of whose zeroes are `-3`and 2, respectively.
Answer» Quadratic equation is of the form=>`x^2-sx+p`, where s and p are sum and product of roots . required equation is=>`x^2+3x+2=0`

Discussion

598.	The graphs of `y=p(x)`are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Answer» Number of zeros for `p(x)` can be finding the number of times a polynomial touches or cuts x-axis. In fig(i), it does not touch x-axis. So, zeros of polynomial is `0`. In fig(ii), it touches x-axis one time. So, zeros of polynomial is `1`. In fig(iii), it touches x-axis `3` times. So, zeros of polynomial is `3`. In fig(iv), it touches x-axis two times. So, zeros of polynomial is `2`. In fig(v), it touches x-axis four times. So, zeros of polynomial is `4`. In fig(vi), it touches x-axis three times. So, zeros of polynomial is `3`.

Discussion

599.	The graphs of y = p(x) are given in the figures below, where p(x) is a polynomial Find the number of zeroes in each case.
Answer» (i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0. (ii) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2. (iii) Here, the graph of y = p(x) intersects the x – axis at four points. So, the number of zeroes is 4. (iv) Here, the graph of y = p(x) does not intersects the x – axis. So, the number of zeroes is 0. (v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2. (vi) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

599.

The graphs of y = p(x) are given in the figures below, where p(x) is a polynomial Find the number of zeroes in each case.

Answer»

(i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0.

(ii) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(iii) Here, the graph of y = p(x) intersects the x – axis at four points. So, the number of zeroes is 4.

(iv) Here, the graph of y = p(x) does not intersects the x – axis. So, the number of zeroes is 0.

(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(vi) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

Discussion

600.	The graphs of y – p(x) are given in the figures below, where p(x) is a polynomial. Find the number of zeros in each case.
Answer» (i) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2. (ii) Here, the graph of y = p(x) intersects the x – axis at three points. So, the number of zeroes is 3. (iii) Here, the graph of y = p(x) intersects the x – axis at one point only. So, the number of zeroes is 1. (iv) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1. (v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2. (vi) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.

600.

The graphs of y – p(x) are given in the figures below, where p(x) is a polynomial. Find the number of zeros in each case.

Answer»

(i) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(ii) Here, the graph of y = p(x) intersects the x – axis at three points. So, the number of zeroes is 3.

(iii) Here, the graph of y = p(x) intersects the x – axis at one point only. So, the number of zeroes is 1.

(iv) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.

(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(vi) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.

Discussion

Explore topic-wise InterviewSolutions in .

Factorise `x^(2)+(1)/(x^(2))-3`.

Factorise `16x^(4)-81y^(4)`.

Factorise `x^(2)-6x-y^(2)+9`.

Factorise `x^(2)-(y-z)^(2)`.

Factorise `a^(2)-9b^(2)`.

Factorise a(x-y)-b(x-y).

Factorise `p^(2)x^(2)+c^(2)x^(2)-ac^(2)-ap^(2)`.

Factorise `x^(2)-(a+(1)/(a))x+1`.

If a + b + c = 0, then a3 + b3 + c3 is equal to (A) 0 (B) abc (C) 3abc (D) 2abc

Write zeros of the polynomial `p(x)=4sqrt(3)x^2+5x-2sqrt(3)`

Factorise ax+ay+az+bx+by+bz.

The coefficient of x in the expansion of (x + 3)3 is (A) 1 (B) 9 (C) 18 (D) 27

If ` p(x)=x^(2)-2sqrt(2x)+1` , then p `(2sqrt(2))` is equal to

the factorisation of `4x^(2)+8x+3` isA. `(x+1)(x+3)`B. `(2x+1)(2x+3)`C. `(2x+2)(2x+5)`D. `(2x-1)(2x-3)`

Which of the following is a factor of (x + y)3 – (x3 + y3 )? (A) x2 + y2 + 2xy (B) x2 + y2 – xy (C) xy2(D) 3xy

the value of `249^(2)-248^(2)` isA. `1^(2)`B. 477C. 487D. 497

The factorisation of 4x2 + 8x + 3 is (A) (x + 1) (x + 3) (B) (2x + 1) (2x + 3) (C) (2x + 2) (2x + 5) (D) (2x –1) (2x –3)

If p(x) = x + 3, then p(x) + p(–x) is equal to(A) 3(B) 2x(C) 0(D) 6

The value of 2492 – 2482 is (A) 12 (B) 477 (C) 487 (D) 497

Zero of the zero polynomial is(A) 0(B) 1(C) Any real number(D) Not defined

One of the factors of (25x2 – 1) + (1 + 5x)2 is (A) 5 + x (B) 5 – x (C) 5x – 1 (D) 10x

If `x+1` is a factor of the polynomial `2x^(2)+kx,` then the value of `k` isA. `-3`B. 4C. 2D. `-2`

x + 1 is a factor of the polynomial(a) x3 + x2 – x +1(b) x3 + x2 + x+1(c) x4 + x3 + x2 +1(d) x4 + 3x3 + 3x2 + x +1

Zero of the polynomial p(x) = 2x + 5 is(A) – 2/5(B) – 5/2(C) 2/5(D) 5/2

If `x^(51) +51` is divided by `x+1` , then the remainder is

One of the zeroes of the polynomial 2x2 + 7x –4 is(A) 2(B) ½(C) – ½(D) –2

If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is (A) –3 (B) 4 (C) 2 (D) –2

If x51 + 51 is divided by x + 1, then the remainder is(a) 0(b)(c) 49(d) 50

Evaluate: (107)2

Evaluate:(94)2

Evaluate the following using suitable identities:(99)3

Evaluate:(0.99)2

Simplify : (3x + 4)3 - (3x - 4)3 .

Evaluate: (1005)3

Evaluate:(997)3

Find the following products of the following expression: (4x + 3y) (16x2 - 12xy + 9y2 )

Find the following products of the following expression:(5x - 2y) (25x2 + 10xy + 4y2 )

Prove that : (x - y)3 + (y - z)3 + (z - x)3 = 3(x - y) (y - z) (z - x).

Find the value of (28)3 - (78)3 + (50)3

If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 - 3abc.

Find a quadratic polynomial with zeroes -2 and 1/3.

Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 + 2x + 1

Find the zeroes of the quadratic polynomial. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial.p(x) = x2 – 4

Find the zeroes of the polynomial `x^2-3`and verify the relationship between the zeroes and the coefficients.

Find a quadratic polynomial, the sum and product of whose zeroes are `-3`and 2, respectively.

The graphs of `y=p(x)`are given in Figure below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

The graphs of y = p(x) are given in the figures below, where p(x) is a polynomial Find the number of zeroes in each case.

The graphs of y – p(x) are given in the figures below, where p(x) is a polynomial. Find the number of zeros in each case.