# Class10 NCRT Polynomials Exercise – 2.4 pdf || UP Board

# __Polynomials__

__Exercise – 2.4__^{*}

__Polynomials__

__Exercise – 2.4__

^{*}**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**

**Solution****:**

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

and the zeroes for p(x) are 1/2 , 1, -2, then putting the zeroes in the equation,

Similarly, putting 1, -2,

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

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

Therefore, all the zeroes of p(x).

Now, comparing with the general polynomial equation,

2x^{3} + x^{2} -5x +2 = ax^{3} + bx^{2} +cx +d

a = 2, b = 1, c = -5, d = 2

Let α=1/2, β=1, γ = -2

½ + 1 + (-2) = -1/2

½ - 1 = -1/2

-1/2 = -1/2

½×1 + 1×-2 + -2×1/2 = -5/2

½ + (-2) + (-1) = -5/2

½ -3 = -5/2

-5/2 = -5/2

½ × 1× (-2) = -2/2

-2/2 = -2/2

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

**(****i****i****)****x ^{3}
- 4x^{2} + 5x - 2; 2 ,1, 1**

**Solution****:**

Let p(x) = x^{3} - 4x^{2} + 5x – 2

and the zeroes for p(x) are 2, 1, 1 then putting the
zeroes in the equation,

p(2) = 2^{3} -4(2)^{2} + 5×2 – 2 = 18
– 18 = 0

Similarly, putting 1, 1,

p(1) = (1)^{3} – 4(1)^{2} + 5(1) - 2 =
0

Therefore, all the zeroes of p(x).

Now, comparing with the general polynomial equation,,

x^{3} - 4x^{2} + 5x – 2 = ax^{3} + bx^{2} +cx +d

a = 1, b = -4, c = 5, d = -2

Let α=2, β=1, γ = 1

2 + 1 + 1 = -(-4/1)

4 = 4

2×1 + 1×1 + 1×2 = 5/1

5 = 5

2 × 1× 1 = -(-2/1)

2 = 2

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
p(x) = ax^{3} + bx^{2} +cx +d be a cubic polynomial and α, β, γ are zeroes

α + β + γ = 2

αβ + βγ + γα = -7

αβγ = -14

We know that,

α + β + γ = -b/a = 2/1

by comparison,

a = 1, b = -2

αβ + βγ + γα = c/a = -7/1

αβγ = -d/a = -14/1

इसी प्रकार c = -7, d = 14

Therefore, cubic polynomial is p(x) = x^{3} - 2x^{2} - 7x +14.

**3. If the zeroes of the polynomial**** ****x ^{3} – 3x^{2} +
x +1**

**are**

**a-b, a, a +b find**

**a**

**and**

**b.**

**Solution****:**

Let α = a-b, β= a, γ= a+b

Sum of zeroes α + β + γ = -b/a

(a-b) + a + (a+b) = -(-3/1)

3a = 3

a = 1

Product of zeroes αβγ = -d/a

(a-b) × a × (a+b) = -(1/1) [a = 1]

1-b^{2} = -1

b^{2} = 2

b = ±√2

Hence, a = 1 and b = ±√2

**4. If two zeroes of the polynomial**** ****x ^{4}-6x^{3}-26x^{2}+13x-35**

**and**

**2±√3**

**find other zeroes.**

**Solution****:**

Let p(x) = x^{4}-6x^{3}-26x^{2}+13x-35

Given that,

2+√3 and 2-√3 are zeroes of the polynomial p(x) so,

x = 2+√3, x = 2-√3

(x-2-√3), (x-2+√3) are factors of the polynomial p(x).

i.e. x^{2}-4x+1 is a factor of the polynomial p(x)

p(x) = x^{4}-6x^{3}-26x^{2}+13x-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-7)(x+5)

Hence x-7=0, x+5=0

The other zeroes of the polynomial are x = 7, –5.

**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****:**

Given that,

Divisor = x^{2}-2x+k

Dividend = x^{4}-6x^{3}+16x^{2}-25x+10

Remainder = x+a

we know that,

Dividend = Divisor × Quotient + Remainder

x^{4}-6x^{3}+16x^{2}-25x+10
= (x^{2}-2x+k) × Quotient + (x + a)

Thus, if the polynomial x^{4}-6x^{3}+16x^{2}-26x+10-a
is divided by the polynomial x^{2}-2x+k, the remainder will be 0.

On comparing,

-10+2k = 0

2k = 10

k = 5

and

10 – a – 8k + k^{2} = 0

10 – a – 8×5 + 5^{2} = 0 [k = 5]

-a-5 = 0

a = -5

Hence, k = 5, a = -5.

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