Explore topic-wise InterviewSolutions in .

Correct option is (C) x³ – 2x² + 4x

Let p(x) be subtracted from

\(x^3 – 2x^2 + 4x +1\) to get 1.

i.e., \(x^3 – 2x^2 + 4x +1-p(x)=1\)

\(\Rightarrow\) \(p(x)=x^3 – 2x^2 + 4x +1--1\)

\(\Rightarrow\) \(p(x)=x^3 – 2x^2 + 4x\)

Correct option is C) x³ – 2x² + 4x

Correct option is (C) \(\frac{4}{x-4}\)

Let p(x) be subtracted from \(\frac{7x}{x^2-x-12}\) to get \(\frac{3}{x+3}\)

i.e., \(\frac{7x}{x^2-x-12}-p(x)=\) \(\frac{3}{x+3}\)

\(\Rightarrow\) \(p(x)=\frac{7x}{(x-4)(x+3)}\) \(-\frac{3}{x+3}\)

\(=\frac{7x-3(x-4)}{(x-4)(x+3)}\) \(=\frac{4x+12}{(x-4)(x+3)}\)

\(=\frac{4(x+3)}{(x-4)(x+3)}\) \(=\frac{4}{x-4}\)

Correct option is C) \(\frac{4}{x-4}\)

p(x) = ax³ + bx² + cx + d

p(l) = a(1)³ + b(1)² + c(1) + d

= a + b + c + d

p(-1) = a(-1)³ + b(-1)² + c(-1) + d

= -a + b – c + d

p(1) = p(-1) is given ie, a + b + c + d = -a + b – c + d

2a + 2c = 0; 2(a + c ) = 0;

ie, a + c = 0

The correct option is: (B) 5

Explanation:

Since, degree of given polynomial is 5, so ax⁵ + bx³ + cx² + dx + e has atmost 5 zeroes.

(a) \(\frac{-b}a\)

Let α , 0 and 0 be the zeroes of ax³ + bx² + cx + d = 0

Then the sum of zeroes = \(\frac{-b}a\)

⇒ α + 0 + 0 = \(\frac{-b}a\)

⇒ α = \(\frac{-b}a\)

Hence, 

the third zero is \(\frac{-b}a\)

Correct option is (D) \(\frac{-x^4 + 31x^2 +20}{x^4 -16}\)

\(\frac{x+2}{x-2}+\frac{x-2}{x+2}-\frac{3(x^2-1)}{x^2+4}\)

\(=\frac{(x+2)^2+(x-2)^2}{(x-2)(x+2)}-\frac{3(x^2-1)}{x^2+4}\)

\(=\frac{2(x^2+4)}{x^2-4}-\frac{3(x^2-1)}{x^2+4}\)

\(=\frac{2(x^2+4)^2-3(x^2-1)(x^2-4)}{(x^2-4)(x^2+4)}\)

\(=\frac{2(x^4+8x^2+16)-3(x^4-5x^2+4)}{x^4-16}\)

\(=\frac{-x^4 + 31x^2 +20}{x^4 -16}\)

Correct option is D) \(\cfrac{-x^4+31x^2+20}{x^4-16}\)

Correct option is (A) 4

Given polynomial is \(7-2x^3+7x^2y+xy^3\)

There is 4 terms in given polynomial.

Now, degree of term 7 is 0.

(constant term / polynomial has degree 0)

Degree of term \(-2x^3\) is 3.

Degree of term \(7x^2y\) is 2+1 = 3.

Degree of term \(xy^3\) is 1+3 = 4

Highest degree among all terms in polynomial is 4.

\(\therefore\) Degree of given polynomial is 4.

Correct option is  A) 4

Given polynomial is 6x²– 3 – 7x 

We have, 6x² – 3 – 7x = 6x² – 7x – 3 

= 6x² – 9x + 2x – 3 

= 3x(2x – 3) + 1(2x – 3) 

= (2x – 3) (3x + 1) 

The value of 6x² – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0 

i.e., when 3x + 1 = 0 and 2x – 3 = 0 

3x = -1 and 2x = 3

x = -1/3 and x = 3/2

∴ The zeroes of 6x² – 3 – 7x = -1/3 and 3/2

∴ Sum of the zeroes = 1/3 + 3/2 = 7/6.

= \(-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-(-7)}{6}=\frac{7}{6}\)

And product of the zeroes = (-1/3) x (3/2) = -1/2

= \(\frac{Constant\,term}{Coefficient\,of\,x^2}= \frac{-3}{6}=\frac{-1}{2}\)

Correct option is C) 2 and 3

(i) 4x² – 3x + 7 

Here polynomial has one variable, i.e. x 

(ii) y² + √2

Here polynomial has one variable, i.e. y 

(iii) 3√t + t√2

This is polynmomial with one variable, because T is only one variable. 

(iv) y + \(\frac{2}{y}\)

Here polynomial has one variable, ie. y. 

(v) x¹⁰ + y³ + t⁵⁰ 

This polynomial is not having one variable because here 3 variables means ‘x’, y and ‘t’ are there.

i. No, because power of v in the term 5√x is -1 (negative number).

ii. No, because the power of x in the term 5√x is i. e. 0.5 (decimal number). 

iii. Yes. All the coefficients are real numbers. Also, the power of each term is a whole number.

iv. No, because the power of m in the term 2m^-2 is -2 (negative number). 

v. Yes, because 10 is a constant polynomial.

2x² +(7/2)x +3/4

The equation can also be written as,

8x²+14x+3

Splitting the middle term, we get,

8x²+12x+2x+3

Taking the common factors out, we get,

4x (2x+3) +1(2x+3)

On grouping, we get,

(4x+1)(2x+3)

So, the zeroes are,

4x+1=0 ⇒ x = -1/4

2x+3=0 ⇒ x = -3/2

Therefore, zeroes are -1/4 and -3/2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

(- 3/2) + (- 1/4) = – (7)/4

= – 7/4 = – 7/4

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 3/2)(- 1/4) = (3/4)/2

3/8 = 3/8

p(x) = kx² − 3x + k

⇒ p(1) = 0

⇒ k(1)² − 3(1) + k = 0

⇒ k − 3 + k = 0

⇒ 2k − 3 = 0

⇒ k = 3/2

Therefore, the value of k is 3/2.

(i) 6x² + 11x + 5 = 0

6x2 + 11x + 5 = 6x2 + 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation 6x² +11x +5 are { −1, −5/6 }

Now, Sum of zeroes of this given polynomial equation = −1+( −56 ) = −11/6

But, the Sum of zeroes of any quadratic polynomial equation is given by = −coeff.of x / coeff.ofx² = −11/6

And Product of these zeroes will be = −1×−5/6 = 5/6

But, the Product of zeroes of any quadratic polynomial equation is given by = constant term / coeff.of x² = 5/6

Hence the relationship is verified.

(ii) 4s2 – 4s + 1

4s2 – 4s + 1 = 4s2 – 2s – 2s + 1

= 2s (2s – 1) –1(2s – 1)

= (2s – 1) (2s – 1)

∴ zeroes of the given polynomial are: {1/2,1/2}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = −coeff. of s / coeff.of s² = −4/4 = 1

And the Product of these zeroes will be = 1/2 × 1/2 =1/4

But, Product of zeroes in any quadratic polynomial equation is given by = constant term/ coeff.of s² = 1/4.

Hence, the relationship is verified.

(iii) 6x2 – 3 – 7x

6x2 – 7x – 3 = 6x2 – 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (−1/3,3/2)

∴ sum of these zeroes will be =  −1/3 + 3/2 =7/6

But, The Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x² = 7/6

And Product of these zeroes will be = −1/3 × 3/2=−1/2
Also, the Product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of x² = −3/6 = −1/2

Hence, the relationship is verified.

(iv) 4u2 + 8u

4u2 + 8u = 4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0

Hence, the zeroes of the above polynomial equation will be (0, −2)

∴ Sum of these zeroes will be = −2

But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.of u / coeff.of u² = −8/4 = −2

And product of these zeroes will be = 0 × −2 = 0

But, the product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of u² = −0/4 = 0

Hence, the relationship is verified.

(v) t² – 15

t² – 15 = (t+ √15) (t − √15)

Therefore, zeroes of the given polynomial are: – {√15, −√15}

∴ sum of these zeroes will be = √15 −  √15 = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x² = −0/1 = 0   

And the product of these zeroes will be = (√15) × (−√15)  = −15

But, the product of zeroes in any quadratic polynomial equation is given by

= constant term / coeff.of t² = −15/1  = −15

Hence, the relationship is verified.

(vi) 3x2 – x – 4

3x2 − x − 4   = 3x2 – 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – {−1, 4/3 }

∴ sum of these zeroes will be = −1 +4/3 = 13

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.of x / coeff.of x²= −(−1)/3 = 1/3

And the Product of these zeroes will be = {−1 × 4/3 }= −4/3

But, the Product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of x² = −4/3

Hence, the relationship is verified.

t³ – 2t² – 15t

Taking t common, we get,

t ( t² -2t -15)

Splitting the middle term of the equation t² -2t -15, we get,

t( t² -5t + 3t -15)

Taking the common factors out, we get,

t (t (t-5) +3(t-5)

On grouping, we get,

t (t+3)(t-5)

So, the zeroes are,

t=0

t+3=0 ⇒ t= -3

t -5=0 ⇒ t=5

Therefore, zeroes are 0, 5 and -3

Verification:

Sum of the zeroes = – (coefficient of x²) ÷ coefficient of x³

α + β + γ = – b/a

(0) + (- 3) + (5) = – (- 2)/1

= 2 = 2

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a

(0)(- 3) + (- 3) (5) + (0) (5) = – 15/1

= – 15 = – 15

Product of all the zeroes = – (constant term) ÷ coefficient of x³

αβγ = – d/a

(0)(- 3)(5) = 0

0 = 0

4x² ˗ 4x + 1 = 0 

⇒ (2x)² ˗ 2(2x)(1) + (1)² = 0

⇒ (2x ˗ 1)² = 0 [∵ a² – 2ab + b² = (a–b)²]

⇒ (2x ˗ 1)² = 0

⇒ x = \(\frac{1}2\) or x = \(\frac{1}2\)

Sum of zeroes = \(\frac{1}2\) + \(\frac{1}2\) = 1 = \(\frac{1}1\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = \(\frac{1}2\) x \(\frac{1}2\) = \(\frac{1}4\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

(i) p(x) = 3x − 2

p(x) = 0

3x − 2 = 0

x = 2/3

Therefore, for x = 2/3,the value of the polynomial is 0 and hence, x = 2/3 is a zero of the given polynomial.

(ii) p(x) = 3x p(x) 

= 0 3x = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

(iii) p(x) = ax p(x) = 0

ax = 0

x = 0

Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.

Solution : Comparing the given polynomial with ax³ + bx² + cx + d, we get
a = 3, b = – 5, c = –11, d = – 3. Further
p(3) = 3 × 3³ – (5 × 3²) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)³ – 5 × (–1)² – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

p(-1/3) = 3* (-1/3)³ - 5*(-1/3)² - 11*(-1/3) - 3,
= -1/9-5/9+11/3-3 = 0

Therefore, 3, -1 and -1/3 are the zeroes of 3x³ - 5x² -11x -3
So, we take α = 3, β = -1, γ = -1/3,

Now,

α + β + γ = 5/3 = -(-5)/3 = -b/a,
αβ + βγ + γα = -11/3 = c/a,
αβγ = 1 = -(-3)/3 = -d/a

f(x) = 2x² ˗ 11x + 15 

= 2x² ˗ (6x + 5x) + 15 

= 2x² ˗ 6x ˗ 5x + 15 

= 2x (x ˗ 3) ˗ 5 (x ˗ 3) 

= (2x ˗ 5) (x ˗ 3) 

∴ f(x) = 0 ⇒ (2x ˗ 5) (x ˗ 3) = 0 

⇒ 2x ˗ 5= 0 or x ˗ 3 = 0

⇒ x = \(\frac{5}2\) or x = 3

So, the zeroes of f(x) are \(\frac{5}2\) and 3.

Sum of zeroes = \(\frac{5}2\) + 3 = \(\frac{5+6}2\) = \(\frac{11}2\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = \(\frac{5}2\) x 3 = \(\frac{-15}2\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)

5t² + 12t + 7

Splitting the middle term, we get,

5t² +5t + 7t + 7

Taking the common factors out, we get,

5t (t+1) +7(t+1)

On grouping, we get,

(t+1)(5t+7)

So, the zeroes are,

t+1=0 ⇒ y= -1

5t+7=0 ⇒ 5t=-7⇒t=-7/5

Therefore, zeroes are (-7/5) and -1

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

(- 1) + (- 7/5) = – (12)/5

= – 12/5 = – 12/5

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 1)(- 7/5) = – 7/5

– 7/5 = – 7/5

We have: 

f(x) = 8x² ˗ 4 

It can be written as 8x² + 0x ˗ 4 

= 4 { (√2x)² ˗ (1)² } 

= 4 (√2x + 1) (√2x ˗ 1) 

∴ f(x) = 0 ⇒ (√2x + 1) (√2x ˗ 1) = 0 

⇒ (√2x + 1) = 0 or √2x ˗ 1 = 0

⇒ x = \(\frac{-1}{\sqrt2}\) or x = \(\frac{1}{\sqrt2}\)

So, the zeroes of f(x) are \(\frac{-1}{\sqrt2}\) and \(\frac{1}{\sqrt2}\)

Here the coefficient of x is 0 and the coefficient of x² is √2

Sum of zeroes = \(\frac{-1}{\sqrt2}\) + \(\frac{1}{\sqrt2}\) = \(\frac{-1+1}{\sqrt2}\) = \(\frac{0}{\sqrt2}\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = \(\frac{-1}{\sqrt2}\) x \(\frac{1}{\sqrt2}\) = \(\frac{-1\times4}{2\times4}\) = \(\frac{-4}{8}\) = \( \frac{constant\,term}{(coefficient\,of\,x^2)}\)

Let α = 2 and β = - 6 

Sum of the zeroes, (α+ β) = 2 + (- 6) = - 4

Product of the zeroes,αβ = 2 × (-6) = -12 

∴ Required polynomial = x² - (α +β)x + αβ = x² – (- 4)x – 12

= x² + 4x – 12

Sum of the zeroes = -4 = \(\frac{-4}1\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = -12 = = \(\frac{-12}1\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)

(i) If x = 1 and x = −1 are zeroes of polynomial p(x) = x² − 1, then p(1) and p(−1) should be 0.

Here, p(1) = (1)² − 1 = 0, and

p(− 1) = (− 1)² − 1 = 0

(ii) If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0.

Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and

p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0

(iii) If x = 0 is a zero of polynomial p(x) = x², then p(0) should be zero. Here, p(0) = (0)² = 0

Hence, x = 0 is a zero of the given polynomial.

(i) If x = -1/√3 and x = 2/√3 are zeroes of polynomial p(x) = 3x² − 1, then

p(-1/√3) and p(2/√3) should be 0.

Here, p(-1/√3) = 3(-1/√3)² - 1

= 1 - 1 = 0,and

p(2/√3) = 3(2/√3)² - 1

= 3(4/3) - 1 = 4 - 1 = 3

Hence, x = -1/√3 is a zero of the given polynomial. However, x = 2/√3 is not a zero of the given polynomial.

(ii) If x = 1/2 is a zero of polynomial p(x) = 2x + 1, then p(1/2) should be .

Here, p(1/2) = 2(1/2) + 1 = 1 + 1 = 2

As p(1/2) ≠ 0

Therefore, x = 1/2 is not a zero of the given polynomial.

3x² + 4x – 4

Splitting the middle term, we get,

3x² + 6x – 2x – 4

Taking the common factors out, we get,

3x(x+2) -2(x+2)

On grouping, we get,

(x+2)(3x-2)

So, the zeroes are,

x+2=0 ⇒ x= -2

3x-2=0⇒ 3x=2⇒x=2/3

Therefore, zeroes are (2/3) and -2

Verification:

Sum of the zeroes = – (coefficient of x) ÷ coefficient of x²

α + β = – b/a

– 2 + (2/3) = – (4)/3

= – 4/3 = – 4/3

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

Product of the zeroes = (- 2) (2/3) = – 4/3

(i) If x = -m/l is a zero of polynomial p(x) = lx + m, then p(-m/l) should be .

Here, p(-m/l) = l(-m/l) + m

= m + m = 0

Therefore, x = -m/l is a zero of the given polynomial.

(ii) If x = 0 is a zero of polynomial p(x) = x², then p(0) should be zero. Here, p(0) = (0)² = 0 

Hence, x = 0 is a zero of the given polynomial.

By splitting the middle term

y² + 3√5/2y - 5 = 0

2y² + 3√5y - 10 = 0

2y² + (4√5y - √5y) - 10 = 0

2y² + (4√5y - √5y) - 10 = 0

2y(y + 2√5) - √5(y + 2√5) = 0

(y + 2√5)(2y - √5) = 0

⇒ y = - 2√5, √5/2

Verification:

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x²

α + β = - b/a

(- 2√5) + (√5/2) = - (3√5)/2

= - 3√5/2 = - 3√5/2

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 2√5)(√5/2) = - 5

- 5 = - 5

Given cubic polynomial 

p(x) = x³ + 3x² – x – 3 

Comparing the given polynomial with ax³ + bx² + cx + d, we get a = 1, b = 3, c = -1, d = -3 

Futher given zeroes are 1,-1 and – 3 

p(1) = (1)³ + 3(1)² – 1 – 3 

= 1 + 3 – 1 – 3 = 0 

p(-1) = (-1)³ + 3(-1)² – 1 – 3 

= -1 + 3 + 1 – 3 = 0 

p(-3) = (-3)³ + 3(-3)² – (-3) – 3 

= -27 + 27 + 3 – 3 = 0

Therefore, 1, -1 and -3 are the zeroes of x³ + 3x² – x – 3. 

So, we take α = 1, β = -1 and γ = -3 

Now, α + β + γ = 1 + (-1) + (-3) = -3

αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1 

= -1 + 3 – 3 

= -1 = c/a = -1/1 = -1 

αβγ = 1 (-1) (-3) = 3 = -d/a = -(-3)/1 = 3

Given polynomial is 3x² – x – 4

we have, 3x² – x – 4 = 3x² + 3x – 4x – 4 

= 3x(x + 1) – 4(x + 1) = (x + 1) (3x – 4) 

The value of 3x² – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0. 

i.e., when x + 1 = 0 or 3x – 4 = 0 

i.e., when x = -1 or x = 4/3

∴ The zeroes of 3x² – x – 4 are -1 and 4/3

Therefore, sum of the zeroes = -1 + 4/3 = -3 +4 /3 = 1/3

= \(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-(-1)}{3}=\frac{1}{3}\)

And product of the zeroes -1 × 4/3 = -4/3

\(=\frac{Constant\,term}{Coefficient\,of\,x^2}=\frac{-4}{3}\)