AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Optional Exercise Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Optional Exercise

### 10th Class Maths 3rd Lesson Polynomials Optional Exercise Textbook Questions and Answers

Question 1.

Verify that the numbers given along-side the cubic polynomials below are their zeroes. Also verify the relation-ship between the zeroes 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 ; (1, 1, 1)

Answer:

i) Given polynomial 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(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,

ii) Given polynomial 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.

Given zeroes are (1, 1, 1)

p(1) = (1)^{3} + 4(1)^{2} + 5(1) – 2

= 1 + 4 + 5 – 2

= 10 – 2 = 8

∴ (1, 1, 1) are not zeroes of the given polynomial p(x).

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 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)

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

a = 1, then 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 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) = \(\frac{-(-3)}{1}\) = 3

So, 3a = 3 ⇒ a = 1

∴ 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 = √2 = ± √2

Here, a = 1 and b = ± √2

Question 4.

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

Answer:

Let the other, two zeroes are α, β.

Then the sum of zeroes of given polynomial = 2 + √3 + 2 – √3 + α + β

= \(\frac{-b}{a}\) = \(\frac{-(-6)}{1}\) = 6

4 + α + β = 6

⇒ α + β = 2 ….. (1)

Now product of zeroes is

(2 + √3 ) (2 – √3) (α) (β)= \(\frac{e}{a}\) = \(\frac{-35}{1}\)

(4 – 3) (αβ) = – 35

⇒ αβ = – 35 …… (2)

Now (α – β)^{2} = (α + β)^{2} – 4αβ

= (2)^{2} – 4(-35) = 4 + 140 = 144

⇒ (α – β) = ± 12 ….. (3)

Now solving (1) & (3) we get

⇒ α = 7; then α + β = 7 + β = 2

⇒ β = -5

The remaining zeroes are α, β

= 7, -5

So total zeroes of given polynomial are 2 + √3, 2 – √3, 7, – 5.

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 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

= x(4k + 25 – 2k – 48) + 10 + k(k + 24)

= x(2k – 23) + (k^{2} + 24k + 10)

Given remainder is x + a

on comparing the coefficients of x and constant terms on both sides

2k – 23 = 1 ……. (1)

2k = 1 + 23 = 24

⇒ k = \(\frac{24}{2}\) = 12

k^{2} + 24k + 10 = a …….. (2)

Substitute ‘k’ value in equation (2)

(12)^{2} + 24(12) + 10 = a

144 + 288 + 10 = a

⇒ a = 442

∴ Required k = 12 and ‘a’ = 442