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27a³ + b³ + 8c³ – 18abc 

= (3a)³ + (b)³ + (2c)³ – 3(3a) (b) (2c) 

= (3a + b + 2c) (9a² + b² + 4c² – 3ab – 2be – 6ca)

a³ – 3a²b + 3ab² – b³ 

= (a)³ – 3 (a)² (b) + 3 (a) (b)² – (b)³ 

= (a – b)³ 

= (a – b) (a – b) (a – b)

4a² + b² + c² – 4ab + 2bc – 4ca 

= (2a)² + (- b)² + (- c)² + 2(2a) (- b) + 2 (- b) (- c) + 2(- c) (2a) 

= (2a – b – c)² = (2a – b – c) (2a – b – c)

(i) 27y³ + 125z³ 

= (3y)³ + (5z)³ 

= (3y + 5z) {(3y)² – (3y)(5z) + (5z)²} 

= (3y + 5z) (9y² – 15yz + 25z²) 

(ii) 64m³ – 343n³ = (4m)³ – (7n)³ 

= (4m – 7n) {(4m)² + (4m) (7n) + (7n)²} 

= (4m – 7n) (17m² + 28mn + 49n²)

Correct option is  C) VII

8a³ – b³ – 12a²b + 6ab² 

= (2a)³ – (b)³ – 3 (2a)² (b) + 3 (2a) (b)² 

= (2a – b)³

1 – 64a³ – 12a + 48a² 

= (1)³ – (4a)³ – 3(1)² (4a) + 3(1) (4a)² 

= (1 – 4a)³

(i) (99)³ = (100 – 1)³ 

As per Identity, 

(a – b)³ = a³ – b³ – 3ab(a – b) 

(100 – 1)³ = (100)³ – (1)³ – 3(100)(1)(100- 1) 

(99)³ = 1000000 – 1 – 300(100 – 1) 

= 1000000 – 1 – 30000 + 300 = 970299 

∴ (99)3 = 970299 

(ii) (102)³ = (100 + 2)³ 

As per Identity,

(a + b)³ = a³ + b³ + 3ab(a + b) 

(100 + 2)³ = (100)³ + (2)³ + 3( 100)(2)( 100 + 2) 

= 1000000 + 8 + 600( 100 + 2) 

= 1000000 + 8 + 60000 + 1200 (100 + 2)³ 

= 1061208 

∴ (102)3= 1061208 

(iii) (998)³ = (1000 – 2)³ 

As per Identity, 

(a – b)³ = a³ – b³ – 3ab(a – b) 

(1000 – 2)³ = (1000)³ – (2)³ – 3( 1000) (2)(1000 – 2) 

= 1000000000 – 8 – 6000(1000 – 1) 

= 1000000000 – 8 – 6000000 + 6000 

= 994005992 

∴ (998)³ = 994005992

Correct option is (A) \(\frac{3abc-b^3}{a^3}\)

\(\alpha+\beta=\frac{-b}a\) and \(\alpha\beta=\frac ca\)

\(\because\) \((\alpha+\beta)^3\) \(=\alpha^3+\beta^3+3\alpha\beta\,(\alpha+\beta)\)

\(=(\frac{-b}a)^3-3.\frac ca.\frac{-b}a\)

\(=\frac{-b^3}{a^3}+\frac{3bc}{a^2}\) \(=\frac{3abc-b^3}{a^3}\)

Correct option is A) \(\cfrac{3abc-b^3}{a^3}\)

Let α and β be the zeroes of the required polynomial f(x). 

Then (α + β) = 8 and αβ = 12 

∴f(x) = x² ˗ (α + β)x + αβ

⇒ f(x) = x² ˗ 8x + 12 

Hence, 

required polynomial f(x) = x² ˗ 8x + 12 

∴f(x) = 0 ⇒ x² ˗ 8x + 12 = 0 

⇒ x² ˗ (6x + 2x) + 12 = 0 

⇒ x² ˗ 6x ˗ 2x + 12 = 0 

⇒ x (x – 6) – 2 (x – 6) = 0 

⇒ (x – 2) (x – 6) = 0 

⇒ (x – 2) = 0 or (x – 6) = 0

⇒ x = 2 or x = 6 

So, the zeroes of f(x) are 2 and 6.

Given : Sum of zeroes = (∝ + β) = 5/2

Product of the zeroes = = 1

Required quadratic polynomial is

x² – (∝+β)x + ∝β 

= x² – (5/2)x + 1

= 1/2(2x² – 5x + 2)

Now, find the zeroes of the above polynomial.

Let f(x) = 1/2(2x² – 5x + 2)

= 1/2(2x² – 4x – x + 2)

= 1/2(2x(x – 2) – (x – 2))

= 1/2( (2x – 1)(x – 2) )

Substitute f(x) = 0.

1/2( (2x – 1)(x – 2) ) = 0

either (2x – 1) = 0 or (x – 2) = o

x = 1/2 or x = 2

1/2 and 2 are the zeroes of the polynomial.

Given : Sum of zeroes = (∝ + β) = 0

Product of the zeroes = = -1

Required quadratic polynomial is

x² – (∝+β)x + ∝β 

= x² – (0)x – 1

= x² – 1

Now, find the zeroes of the above polynomial.

Let f(x) = x² – 1

= x² – 1²

= (x – 1)(x + 1)

Substitute f(x) = 0.

either (x – 1) = 0 or (x + 1) = o

x = 1 or x = -1

1 and -1 are the zeroes of the polynomial.

Solution:

According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x)
Degree of a polynomial is the highest power of the variable in the polynomial.
(i) deg p(x) = deg q(x)
Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).
Let us assume the division of 6x² + 2x + 2 by 2.
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1and r(x) = 0
Degree of p(x) and q(x) is the same i.e., 2.
Checking for division algorithm, p(x) = g(x) × q(x) + r(x)
6x² + 2x + 2 = (2) (3x² + x + 1) + 0
Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)
Let us assume the division of x³ + x by x²,
Here, p(x) = x³ + x g(x) = x² q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1. Checking for division algorithm, p(x) = g(x) × q(x) + r(x)

x³ + x = (x² ) × x + x x³ + x = x³ + x
Thus, the division algorithm is satisfied.

(iii) deg r(x) = 0
Degree of remainder will be 0 when remainder comes to a constant.
Let us assume the division of x³ + 1 by x².
Here, p(x) = x³ + 1 g(x) = x² q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0. Checking for division algorithm,
p(x) = g(x) × q(x) + r(x) x³ + 1 = (x² ) × x + 1 x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.

Solution:

mx²+(m+n)x+n

=mx²+mx+nx+n

=mx(x+1)+n(x+1)

=(mx+n)(x+1)

So, x = -n/m or x = -1

two variables binomial

x³ y² + xy

Number of trees in the village Lat = a 

Number of trees increasing each year = b 

∴ Number of trees after x years = a + bx 

∴ There will be a + bx trees in the village Lat after x years.

Total rows = x 

Number of students in each row = y 

∴ Total students = Total rows × Number of students in each row.

= x × y 

= xy 

∴ There are in all xy students for the parade.

Digit in units place = n 

Digit in tens place = m 

∴ The two digit number = 10 x digit in tens place + digit in units place

 = 10m + n

∴ The polynomial representing the two digit number is 10m + n

Correct option is C) Multinomial

Correct option is D) 1

Correct option is D) 3

Correct option is B) 3

Correct option is (B) 2x² – 5 = 0

Let breadth of rectangular dining hall be x unit.

Then length of rectangular dining hall = 2x unit.

\(\therefore\) Area of rectangular dining hall \(=lb=2x^2\) sq. units

\(\therefore\) \(2x^2\) = 5   (Given area = 5 sq. units)

\(\Rightarrow\) \(2x^2\) - 5 = 0     _______(1)

Equation (1) is the polynomial equation which represents the situation.

Correct option is B) 2x² – 5 = 0

Using the identity : (x + a) (x + b) = x² + (a + b) x + ab, we have 

(x + 2) (x + 3) = x² + (2 + 3) x + 2 × 3 

= x² + 5x + 6. 

Using the identity : (x + a) (x + b) = x² + (a + b) x + ab, we have 

(x + 7) (x - 2) = (x + 7) (x + (-2)) 

 = x² + 7x + (-2)x + 7 × (-2) 

 = x² + 5x - 14

Using the identity : (x + a) (x + b) = x² + (a + b) x + ab, we have 

(y - 4) (y- 3) = {y + (-4)} {y+ (-3)} 

 = y² + {(-4) + (-3)}y + (-4) × (-3) 

 = y² - 7y + 12 

Using the identity : (x + a) (x + b) = x² + (a + b) x + ab, we have

 (y - 7) (y + 3) = {y + (-7)} (y + 3) 

 = y² + {(-7) + 3} + (-7) × 3 

 = y² - 4y - 21. 

Using the identity : (x + a) (x + b) = x² + (a + b) x + ab, we have 

(2x - 3) (2x + 5) = (y - 3) (y + 5), where y = 2x 

 = {y + (-3)} (y + 5) 

 = y² + {(-3) + 5} y + (-3) × 5 

 = y² + 2y - 15 

 = (2x)² + 2 × 2x - 15 

 = 4x² + 4x - 15

(3x + 4) (3x - 5) = (y + 4) (y - 5), where y = 3x 

 = (y+ 4) {y + (-5)} 

 = y² + {4 + (-5)} + 4 × (-5) 

 = y² - y - 20 

 = (3x)² - 3x - 20 

 = 9x² - 3x - 20

35 × 37 = (40 - 5) (40 - 3) 

= (40 + (-5)) (40 + (-3)) 

= 40² + (-5 - 3) × 40 + (-5 × - 3) 

= 1600 - 320 + 15 

= 1615 - 320 

= 1295 

103 × 96 

= (100 + 3) [100 + (-4)] 

= 100² + (3 + (-4)) × 100 + (3 × - 4) 

= 10000 - 100 - 12 

= 9888 

81a²b 2bc² + 64a⁶b ² - 144a⁴b ² c 

= [9abc]² -2 [9abc][8a³b] + [8a³b]² 

= [9abc - 8a³b]² 

 = a²b ² [9c - 8a² ] ² 

Answer is (D) 3

Putting x = 1 in equation

ax² + ax + 3 = 0

⇒ a(1)² + a(1) + 3 = 0

⇒ a + a + 3 = 0

⇒ 2a = – 3

⇒ a = -3/2

and putting x = 1 in equation

x² + x + b = 0

⇒ (1)² + (1) + b = 0

⇒ 1 + 1 + b = 0

⇒ b = -2

ab = -3/2 × -2 = 3

Answer is (B) 2, - 3/2

Given quadratic equation is :

2x² – x – 6 = 0

⇒ 2x² – (4 – 3) x – 6 = 0

⇒ 2x² – 4x + 3x – 6 = 0

⇒ (2x² – 4x) + (3x – 6) = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ (x – 2)(2x + 3) = 0

⇒ x-2 = 0 or 2x + 3 = 0

⇒ x = 2 or x = -3/2

Answer is (B) -x² + 3x – 3 = 0

Sum of roots of option (A) = -b/a = 3/2

Sum of roots of option (B) = -b/a = 3

Answer is (D) ±3

x² – 9 = 0

⇒ x² = 9

⇒ x = ±√9

⇒ x = ± 3

Answer is (D) 0, 8

Answer is (C) 4

Given equation x² – kx + 4 = 0

Comparing it with ax² + bx + c = 0

We get a = 1, b= -k and c = 4

D = b² – 4 ac = (-k)² – 4 × 1 × 4 = k² – 16

Roots are same

D = 0

⇒ k² – 16 = 0

⇒ k² = 16

⇒ k = ±√16

⇒ k = ± 4

Answer is (A) -3

Equation 2x² + x – 6 = 0

Here a = 2, b = 1, c = -6

Product of root = c/a = -6/2 = -3

Answer is (C) No real root

Given

(k – 12)x² + 2(k – 12)x + 2 = 0

Comparing it with quadratic equation 

ax² + bx + c = 0

a = (k – 12),b = 2(k – 12), c = 2

Discriminant (D) = b² – 4ac

= 4(k – 12)² – 4 × (k – 12) × 2

= 4(k – 12) [k – 12 – 2]

= 4(k – 12)(k – 14)

Given equation will have real and equal roots, if discriminant = 0.

D = 0

⇒ 4(k – 12)(k – 14) =0

⇒ k – 12 = 0 or k – 14 = 0

⇒ k = 12 or k = 14

(i) Given roots kx² + 2x + 1 = 0

Comparing it by general quadratic equation

ax² + bx + c = 0

a = k, b = 2, c = 1

Discriminant (D) = b² – 4ac

= (2)² – 4 × k × 1

= 4 – 4 k

= 4(1 – k)

For real and fractional roots

Discriminant (D) > 0

⇒ 4(1 -k) > 0

⇒ 1 – k > 0

⇒ k < 1

Thus, k will be less than.

(ii) Given equation kx² + 6x + 1 = 0

Comparing it by general quadratic equation ax² + bx + c – 0,

a = k, b = 6, c = 1

Discriminant (D) = b² – 4ac

= (6)² – 4 × k × 1

= 36 – 4k

= 4(9 – k)

If discriminant (D) > 0 then roots of equation will be real and fractional.

i.e., D > 0

⇒ 4(9 – k) > 0

⇒ 9 – k > 0

⇒ k > 9

Thus value of k will be greater than 9

(iii) Given equation x² – kx + 9 = 0

Comparing it by general quadratic equation

ax² + bx + c = 0

a = 1, b = -k, c = 9

Discriminant (D) = b² – 4ac

= (-k)² – 4 × 1 × 9

= k² – 36

If D > 0, then roots of equation will be real and fractional.

D > 0

⇒ k² – 36 > 0

⇒ (k – 6)(k + 6) > 0

⇒ k – 6 > 0 or k + 6 > 0

⇒ k > 6 or k < -6

Thus, value of k will be greater than 6 or less than -6

Given equation x² + 5kx + 16 = 0

Comparing it by general quadratic equation ax² + bx + c = 0

a = 1, b = 5k, c = 16

Discriminant (D) = b² – 4ac

= (5k)² – 4 × 1 × 16

= 25k² – 64

If Discriminant (D) < 0, then roots will not be real.

i.e. D < 0

⇒ 25k² – 64 < 0

⇒ (5k – 8)(5k + 8) < 0

⇒ 5k – 8 < 0 or 5k + 8 < 0

⇒ k < 8/5 or k > -8/5

Hence, value of k will be smaller than 8/5 or greater than -8/5

(1001)³ = (1000 + 1)³. 

[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³] 

= 1000³ + 3(1000)² (1) + 3(1000) (1)² + 1³

= 1000000000 + 3000000 + 3000 + 1 

= 1003003001

Class 10 Maths MCQ Questions of Polynomials with Answers was Prepared Based on Latest Pattern of exams. Students can solve MCQ Questions with Answers to realize their preparation level. Practice more and more these Multiple Questions will assist understudies with scoring better stamps in the board tests.

Check out Multiple Choice Questions of Polynomials are given underneath: -

                           Practice Class 10 Maths MCQ Question of Polynomials

1. If a polynomial p(y) is divided by y + 2, then which of the following can be the remainder:

(a)y + 1
(b)2y + 3
(c) 5
(d)y – 1

2. If the polynomials \(ax^3 + 4x^2 + 3x – 4\) and \(x^3 – 4x + a\), leave the same remainder when divided by \((x – 3)\), then value of a is :

(a) 2b
(b) – 1
(c) 1
(d) – 2b

3. If -2 is a zero of \(p(x) = (ax³ + bx² + x – 6)\) and \(p(x)\) leaves a remainder 4 when divided by \((x – 2)\), then the values of a and b are (respectively):

(a) a = 2,b = 2
(b) a = 0,b = – 2
(c) a = 0, b = 2
(d) a = 0, b = 0

4. If \(x^{101} + 1001\) is divided by \(x + 1\), then remainder is:

(a) 0
(b) 1
(c) 1490
(d) 1000

5. The degree of the polynomial \((x + 1)(x^2 – x – x^4 +1)\) is:

(a) 2
(b) 3
(c) 4
(d) 5

6. Find the remainder when \(x^4+x^3-2x^2+x+1\) is divided by \(x-1\)

(a) 1
(b) 5
(c) 2
(d) 3

7. If a and b are the zeroes of the polynomial \(x^2-11x +30\), Find the value of \(a^3 + b^3\)

(a) 134
(b) 412
(c) 256
(d) 341

8. \(S(x) = px^2+(p-2)x +2\). If 2 is the zero of this polynomial,what is the value of p

(a) -1
(b) 1/2
(c) -1/2
(d) +1

9. If the zeroes of the quadratic polynomial \(ax^2 + bx + c, c ≠ 0\) are equal, then

(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign

10. What are the zeroes of the Polynomial \(p(x)=x^2+7x+10\)

(a) (2) , (-5)
(b) (-2) , (5)
(c) (2) , (5)
(d) (-2) , (-5)

11. The graph of the polynomial \(ax^2 + bx + c\) is an upward parabola if

(a) a > 0
(b) a < 0
(b) a = 0
(d) None

12. A polynomial of degree 3 is called

(a) a linear polynomial
(b) a quadratic polynomial
(c) a cubic polynomial
(d) a biquadratic polynomial

13. Dividend is equal to

(a) divisor × quotient + remainder
(b) divisior × quotient
(c) divisior × quotient – remainder
(d) divisor × quotient × remainder

14. The graph of the polynomial \(ax^2 + bx + c\) is a downward parabola if

(a) a > 0
(b) a < 0
(c) a = 0
(d) a = 1

15. If α, β are the zeros of the polynomial \(f(x) = x^2 + x + 1\), then \(\frac{1}{\alpha}+\frac{1}{\beta}=\)

(a) 1
(b) – 1
(c) 0
(d) None of these

16. If one zero of the polynomial \(f(x) = (k^2 + 4) x^2 + 13x + 4k\) is reciprocal of the other, then k =

(a) 2
(b) – 2
(c) 1
(d) – 1

17. If the sum of the zeros of the polynomial \(f(x) = 2x^3 – 3kx^2 + 4x – 5\) is 6, then the value of k is

(a) 2
(b) 4
(C) – 2
(d) – 4

18. sum of the squares of the zeroes of the polynomial \(p(x) = x^2 + 7x – k\) is 25, find k.

(a) 12
(b) 49
(c) – 24
(d) – 12

19. Let, α, s, v be the zeroes of \(x^3 + 4x^2 + x- 6\) such that the product of two of the zeroes is 6. Find the third zero.

(a) 6
(b) 2
(c) 4
(d) 1

20. Find a and b so that the polynomial \(6x^4 + 8x^3 – 5x^2 + ax + b\) is exactly divisible by \(2x^2 – 5\).

(a) a = 20, b = – 25
(b) a = 4, b = – 5
(c) a = 20, b = 5
(d) a = – 20, b = – 25

Answers & Explanations

1. Answer: (c) 5

Explanation: When p(y) is divided by y + 2, then the degree of remainder < deg of (y + 2)

2. Answer: (b) – 1

Explanation: \(p(x) = ax^3 + 4x^2 + 3x – 4\)
\(q(x) = x^3 – 4x + a\)
p(3) = q(3)
a = – 1

3. Answer: (c) a = 0, b = 2

Explanation: If – 2 is a zero
\(\Rightarrow\)p(- 2) = 0
\(\Rightarrow\) – 2a + b = 2
Also, p(2) = 4
2a + b = 2
\(\Rightarrow\)a = 0 and b = 2

4. Answer: (d) 1000

Explanation: \(p(x)\) is divided by \(x + 1\)
\(p(- 1) = (-1^{101}) + 1001 = 1000\)

5. Answer: (d) 5

Explanation: Since the highest degree variable in first bracket is \(x\) and in second bracket is \(x^4\) on multiplying \(x\) with \(x^4\).the highest power we obtain is 5.

6. Answer: (c) 2

Explanation: Let \(P(x) =x^4+x^3-2x^2+x+1\)
Remainder when divide by \(x-1\)
\(P(1) = 1+ 1-2+1+1=2\)

7. Answer: (d) 341

Explanation: \(a^3 + b^3= (a+b) (a^2+b^2-ab)=(a+b) \{(a+b)^2 -3ab\}\)
Now a+b=-(-11)/1=11
ab=30
So \(a^3+b^3=11( 121 -90)=341\)

8. Answer: (c) -1/2

Explanation: \(S(2)=4p+0+2=0 \Rightarrow p=-1/2\)

9. Answer: (c) c and a have the same sign

Explanation: Given that, the zeroes of the quadratic polynomial \(ax^2 + bx + c , c ≠ 0\) are equal
Value of discriminant (D) has to be zero
\(b^2– 4ac = 0\)
\(b^2 = 4ac\)
Since L.H.S. \((b^2)\) can never be negative
R.H.S. also can never be negative.
a and c must have same sign.

10. Answer: (d) (-2) , (-5)

Explanation: \(p(x)=x^2+7x+10\)
\(\Rightarrow p(x) = x^2 + 5x +2x +10 = x(x +5) + 2(x+5)=(x+2)(x+5)\)
So roots are (-2) , (-5)

11. Answer: (a) a > 0

12. Answer: (c) a cubic polynomial

13. Answer: (a) divisor × quotient + remainder

14. Answer: (b) a < 0

15. Answer: (b) – 1

Explanation: Given,
\(f(x) = x^2 + x + 1\)
α = a
β = b
c = c
α + β = – b/a
= – 1/1
= – 1 …….. Equ.(i)
αβ = c/a = 1 ….. Equ.(ii)
Dividing (i) from (ii)
We get,
– 1/1 = – 1
So,
1/α + 1/β = – 1

16. Answer: (a) 2

Explanation: Given;

f(x) = (k² + 4) x² + 13x + 4k,

One zero of the polynomial is reciprocal of the other,

Let a be the one zero,

∴ The other zero will be 1/a

As we know that,

Product of the zeros = c/a = 4k/k² + 4

∴ 4k/k² + 4 = 1

⇒ 4k = k² + 4

⇒ k ² + 4 – 4k = 0

⇒ (k – 2)² = 0

⇒ k = 2

So the value of k is 2

17. Answer: (b) 4

Explanation: Given,

f(x) = 2x³ – 3kx² + 4x – 5

Sum of the zeros of the polynomial = 6

Let x, y and z be the zeroes than,

x + y + z = 6……Equation (i)

So,

x + y + z = – b/a

= – ( – 3k)/2

From Eq. (i) we get,

3k/2 = 6

k/2 = 6/3

k = 2×2 = 4

18. Answer: (d) – 12

Explanation: \(p(x) = x^2 + 7x – k\)
let α, s be the zeroes
α + s = – 7
αs = – k
\(α^2 + s^2 = 25\)
\((α^2 + s) – 2αs = 25\)
49 + 2k = 25
k = -12

19. Answer: (d) 1

Explanation: α s v = 6,
αs = 6(given)
\(\Rightarrow v = 6/6=1\)

20. Answer: (d) a = – 20, b = – 25

Explanation: Divide the given polynomial by \(2×2 – 5\) get the remainder as \((20 + a)x + (b + 25)\) which should be zero.

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