TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise

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) 2x³ + x² – 5x + 2;(\(\frac{1}{2}\), 1, -2)
ii) x³ + 4x² + 5x – 2; (2, 1, 1)
Solution:
i) Given polynomial is 2x³ + x² – 5x + 2
Comparing the given polynomial with ax³ + bx² + 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 2x³ + x² – 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 x³ – 4x² + 5x – 2
Comparing the given polynomial with ax³ + bx² + 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 x³ – 4x² + 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
ax³ + bx² + 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
x³ – 2x² – 7x + 14

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Answer:
Given polynomial is x³ – 3x² + x + 1
Since, (a – b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + 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
⇒ a² – ab + a² + ab + a² – b² = 1
⇒ 3a² – b² = 1
So, 3(1)² – b² = 1
⇒ 3 – b² = 1
⇒ b² = 2 ⇒ b = \(\sqrt{2}\) = ±\(\sqrt{2}\)
Here, a = 1 and b = ±\(\sqrt{2}\)

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\) , find other zeroes.
Answer:
Given polynomial is
x⁴ – 6x³ – 26x² + 138x – 35
We have, 2 ± \(\sqrt{3}\) are two zeroes of the polynomial
p(x) = x⁴ – 6x³ – 26x² + 138x – 35
Let x = 2 ± \(\sqrt{3}\)
So, x – 2 = ±\(\sqrt{3}\)
On squaring, we get x² – 4x + 4 = 3, i.e., x² – 4x + 1 = 0
Let us divide p(x) by x² – 4x + 1 to obtain other zeroes

p(x) = x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) (x² – 7x + 5x – 35)
= (x² – 4x + 1) [x(x – 7) +5(x – 7)]
= (x² – 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
x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Answer:
Given polynomial is
x⁴ – 6x³ + 16x² – 25x + 10 and another polynomial is x² – 2x + k
Remainder is x + a
Let us divide
x⁴ – 6x³ + 16x² – 25x + 10 by x² – 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