Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise

Question 1.

Verify that the numbers given along side the cubic polynomials below are their zeroes. Also verify the relationship between the zeros and the coefficients in each case :

i) 2x^{3} + x^{2} – 5x + 2;(\(\frac{1}{2}\), 1, -2)

ii) x^{3} + 4x^{2} + 5x – 2; (2, 1, 1)

Solution:

i) Given polynomial is 2x^{3} + x^{2} – 5x + 2

Comparing the given polynomial with ax^{3} + bx^{2} + cx + d,

We get a = 2, b = 1, c = -5 and d = 2

p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))^{3} + (\(\frac{1}{2}\))^{3} – 5(\(\frac{1}{2}\)) + 2

= \(\frac{1}{4}\) + \(\frac{1}{4}\) – \(\frac{5}{2}\) + \(\frac{2}{1}\)

= \(\frac{1+1-10+8}{4}\) = \(\frac{0}{4}\) = 0

P(1) = 2(1)^{3} + (1)^{2} – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

p(-2) = 2(-2)^{3} + (-2)^{2} – 5(-2) + 2

= 2(-8) + 4 + 10 + 2

= -16 + 16 = 0

∴ \(\frac{1}{2}\), 1, and -2 are the zeroes of 2x^{3} + x^{2} – 5x + 2

So, α = \(\frac{1}{2}\), β = 1 and γ = -2

Therefore,

α + β + γ = \(\frac{1}{2}\) + 1 + (-2)

= \(\frac{1+2-4}{2}\) = \(\frac{-1}{2}\) = \(\frac{-b}{a}\)

αβ + βγ + γα = (\(\frac{1}{2}\))(1) + (1)(-2) + -2(\(\frac{1}{2}\))

= \(\frac{1}{2}\) – 2 – 1

= \(\frac{1-4-2}{2}\) = \(\frac{-5}{2}\) = \(\frac{c}{a}\)

And αβγ = \(\frac{1}{2}\) × 1 × (-2) = -1 = \(\frac{-2}{2}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

ii) Given polynomial is x^{3} – 4x^{2} + 5x – 2

Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, We get a = 1, b = -4, c = 5 and d = -2

p(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2 = 0

p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

∴ 2, 1 and 1 are the zeroes of x^{3} – 4x^{2} + 5x – 2.

So, α = 2, β = 1 and γ = 1

Therefore,

α + β + γ = 2 + 1 + 1

= 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\)

αβ + βγ + γα = 2(1) + (1)(1) + (1) (2)

= 2 + 1 + 2 = 5

= \(\frac{5}{1}\) = \(\frac{c}{a}\)

and

αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Question 2.

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes as 2, -7, -14 respectively.

Answer:

Let the cubic polynomial be

ax^{3} + bx^{2} + cx + d and its zeroes be α, β and γ.

Then

α + β + γ = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-b}{a}\)

αβ + βγ + γα = -7 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)

and αβγ = -14 = \(\frac{-14}{1}\) = \(\frac{-d}{a}\)

∴ a = 1, b = -2, c = -7 and d = 14.

So, one cubic polynomial which satisfies the given conditions will be

x^{3} – 2x^{2} – 7x + 14

Question 3.

If the zeroes of the polynomial x^{3} – 3x^{2} + x + 1 are a – b, a, a + b, find a and b.

Answer:

Given polynomial is x^{3} – 3x^{2} + x + 1

Since, (a – b), a, (a + b) are the zeroes of the polynomial x^{3} – 3x^{2} + x + 1.

Therefore, sum of the zeroes = (a – b) + a + (a + b)

⇒ 3a = 3 ⇒ a = 1

= \(\frac{-(-3)}{1}\) = 3

∴ Sum of the products of its zeroes taken two at a time.

= a(a – b) +a(a + b) + (a + b) (a – b)

= \(\frac{1}{1}\) = 1

⇒ a^{2} – ab + a^{2} + ab + a^{2} – b^{2} = 1

⇒ 3a^{2} – b^{2} = 1

So, 3(1)^{2} – b^{2} = 1

⇒ 3 – b^{2} = 1

⇒ b^{2} = 2 ⇒ b = \(\sqrt{2}\) = ±\(\sqrt{2}\)

Here, a = 1 and b = ±\(\sqrt{2}\)

Question 4.

If two zeroes of the polynomial x^{4} – 6x^{3} – 26x^{2} + 138x – 35 are 2 ± \(\sqrt{3}\) , find other zeroes.

Answer:

Given polynomial is

x^{4} – 6x^{3} – 26x^{2} + 138x – 35

We have, 2 ± \(\sqrt{3}\) are two zeroes of the polynomial

p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35

Let x = 2 ± \(\sqrt{3}\)

So, x – 2 = ±\(\sqrt{3}\)

On squaring, we get x^{2} – 4x + 4 = 3, i.e., x^{2} – 4x + 1 = 0

Let us divide p(x) by x^{2} – 4x + 1 to obtain other zeroes

p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35

= (x^{2} – 4x + 1) (x^{2} – 2x – 35)

= (x^{2} – 4x + 1) (x^{2} – 7x + 5x – 35)

= (x^{2} – 4x + 1) [x(x – 7) +5(x – 7)]

= (x^{2} – 4x + 1) (x + 5) (x – 7)

So, (x + 5) and (x – 7) are other factors of p(x)

So, -5 and 7 are other zeroes of the given polynomial.

Question 5.

If the polynomial

x^{4} – 6x^{3} – 16x^{2} + 25x + 10 is divided by another polynomial x^{2} – 2x + k, the remainder comes out to be x + a, find k and a.

Answer:

Given polynomial is

x^{4} – 6x^{3} + 16x^{2} – 25x + 10 and another polynomial is x^{2} – 2x + k

Remainder is x + a

Let us divide

x^{4} – 6x^{3} + 16x^{2} – 25x + 10 by x^{2} – 2x + k.

∴ Remainder = (2k – 9)x – (8 – k)k + 10

But the remainder is given as x + a.

On comparing their coefficients we have

2k – 9 = 1 ⇒ 2k = 10

⇒ k = 5 and -(8 – k) k + 10 = a

So, a = -(8 – 5) 5 + 10 = -3 × 5 + 10

= -15 + 10 = -5

Hence, k = 5 and a = -5