**1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:**

**(i) 2x ^{3 }+ x^{2} – 5x + 2; ^{1}/_{2} , 1, -2**

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

Solution:

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

p(x) = 2x^{3 }+ x^{2} – 5x + 2

Zeros for this polynomial are ^{1}/_{2 }, 1, -2

p(^{1}/_{2}) = 2(^{1}/_{2})^{3 }+ (^{1}/_{2})^{2} – 5(^{1}/_{2}) + 2

= (^{1}/_{4}) + (^{1}/_{4}) – (^{5}/_{2}) + 2

= 0

p(1) = 2(1)^{3 }+ (1)^{2} – 5(1) + 2 = 0

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

= -16 + 4 + 10 + 2

= 0

Therefore, ½ , 1, and −2 are the zeroes of the given polynomial. Comparing the given polynomial with ax^{3} + bx^{2 }+ cx + d, we obtain a = 2, b = 1, c = −5, d = 2

Let us take α = ^{1}/_{2} ,β = 1, γ = -2

α + β + γ = ^{1}/_{2} + 1 + (-2) = –^{1}/_{2} = –^{b}/_{a}

αβ + βγ + γα = ^{1}/_{2} x 1 + 1(-2) + ^{1}/_{2} (-2) = ^{-5}/_{2 }= ^{c}/_{a}

αβγ = ^{1}/_{2} x 1 x (-2) = –^{1}/_{1 }= – ^{2}/_{2 }= –^{d}/_{a}

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii) x^{3} – 4x^{2} + 5x – 2; 2,1,1

p(x) = x^{3 }– 4x^{2} + 5x – 2

Zeros for this polynomial are 2 , 1, 1

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 = 0

Therefore, 2 , 1, and 1 are the zeroes of the given polynomial. Comparing the given polynomial with ax^{3} + bx^{2 }+ cx + d, we obtain a = 1, b = -4, c = 5, d = -2

Verify:

Sum of zeros = 2+1+1 = 4 = – ^{(-4)}/_{1} = –^{b}/_{a}

Multiplication of zeroes taking two at a time = (2)(1)+(1)(1)+(2)(1) = 2+1+2= 5 = ^{5}/_{1} = ^{c}/_{a}

Multiplication of zeroses = 2 x 1 x 1 = 2 = – ^{(-2)}/_{1} = –^{d}/_{a}

Hence, the relationship between the zeroes and the coefficients is verified.

**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. **

Solution:

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

It is given that

α + β + γ = ^{2}/_{1} = –^{b}/_{a}

αβ + βγ + γα = ^{-7}/_{2 }= ^{c}/_{a}

αβγ = – ^{14}/_{1 }= –^{d}/_{a}

If a = 1, then b = −2, c = −7, d = 14 Hence, the polynomial is x^{3} – 2x^{2} – 7x + 14

**3. ****If the zeroes of polynomial, x ^{3} -3x^{2}+x+1 are a-b,a,a+b. Find a and b.**

Solution:

p(x) = x^{3} -3x^{2}+x+1

Zeroes are a − b, a + a + b Comparing the given polynomial with px^{3}+qx^{2}+rx+1, we obtain p = 1, q = −3, r = 1, t = 1

Sum of zeros = a – b + a + a + b

^{-q}/_{ p} = 3a

–^{(-3)}/_{1 } = 3a

3 = 3a

a = 1

The zeroes are 1-b, 1, 1+b

Multiplication of zeroes = 1(1-b)(1+b)

^{-t}/_{p} = 1 – b^{2}

^{-1}/_{1}= 1 – b^{2}

1 – b^{2} = -1

1 + 1 = b^{2}

b = ±√2

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

Solution:

Given 2 + √3 and 2 – √3 are zeroes of the given polynomial. So, (2 + √3)(2 – √3) is a factor of polynomial.

Therefore,[x – (2 + √3)][x – (2 – √3)] = x^{2} + 4 − 4x − 3 = x^{2} − 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find

Clearly, x^{4} -6x^{3} -26x^{2}+138x -35 = (x^{2} – 4x + 1)(x^{2} – 2x – 35)

We know, (x^{2} – 2x – 35) = (x – 7)(x + 5)

Therefore, the value of the polynomial is also zero when (x -7 = 0) or (x + 5 = 0)

Or x = 7 or −5

therefore, 7 and −5 are also zeroes of this polynomial.

**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.**

Solution:

By division algorithm we know,

Dividend = Divisor × Quotient + Remainder

Dividend − Remainder = Divisor × Quotient

x^{4} – 6x^{3} + 16x^{2} – 25x + 10 – x – a = x^{4} – 6x^{3} + 16x^{2} – 26x + 10 – a will be divisible by x^{2} – 2x + k

It ccan be observed that (-10+2k)x +(10-a-8k+k^{2}) will be zero.

Thus, (-10 + 2k) = 0 and (10 – a – 8k +k^{2}) = 0

For (-10+2k) = 0 , 2k = 10 and thus k = 5

For (10 – a – 8k + k^{2}) = 0

10 − a − 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0

− 5 − a = 0

Therefore, a = −5

Hence, k = 5 and a = −5

## 3 responses to “Polynomials – Exercise 2.4 – Class 10”

[…] Polynomials – Exercise 2.4 […]

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