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

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

**Answer
1** :

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

**Solution:**

Given, p(x) **= **2x^{3}+x^{2}-5x+2

And zeroes for p(x) 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

= 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x^{3}+x^{2}-5x+2.

Now, comparing the given polynomial withgeneral expression, we get;

∴ ax^{3}+bx^{2}+cx+d

= 2x^{3}+x^{2}-5x+2

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

As we know, if α, β, γ are the zeroes of thecubic polynomial ax^{3}+bx^{2}+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of thepolynomial,

α+β+γ = ½+1+(-2)

= -1/2

= –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2)

= -5/2

= c/a

α β γ = ½×1×(-2)

= -2/2

= -d/a

Hence, the relationship between the zeroes andthe coefficients are satisfied.

**(ii) x ^{3}-4x^{2}+5x-2** ;

**Solution:**

Given, p(x) = x^{3}-4x^{2}+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 2^{3}-4(2)^{2}+5(2)-2

= 0

p(1) = 1^{3}-(4×1^{2 })+(5×1)-2

= 0

Hence proved, 2, 1, 1 are the zeroes of x^{3}-4x^{2}+5x-2

Now, comparing the given polynomial withgeneral expression, we get;

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

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

As we know, if α, β, γ are the zeroes of thecubic polynomial ax^{3}+bx^{2}+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of thepolynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes andthe coefficients are satisfied.

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

**Answer
2** :

**Solution:**

We are given with the polynomial here,

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

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial withgeneral expression, we get;

∴px^{3}+qx^{2}+rx+s = x^{3}-3x^{2}+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b^{2}

-1/1 = 1-b^{2}

b^{2} = 1+1 = 2

b = √2

Hence,1-√2, 1 ,1+√2 are the zeroes of x^{3}-3x^{2}+x+1.

**Answer
3** :

**_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
4** :

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