**Polynomials**

**Exercise 2.4 Page: 36**

**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:**

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 with general 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 the cubic 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 the polynomial,

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 and the coefficients are satisfied.

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

**2, 1, 1**

**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 with general 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 the cubic 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 the polynomial,

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 and the coefficients are satisfied.

**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 us consider the cubic polynomial is ax^{3}+bx^{2}+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is 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:**

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 with general 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.

**4. If two zeroes of the polynomial x ^{4}-6x^{3}-26x^{2}+138x-35**

**are 2 ±**√

**3,**

**find other zeroes.**

**Solution:**

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x^{4}-6x^{3}-26x^{2}+138x-35

Since 2 +√**3 **and 2-√**3 **are zeroes of given polynomial f(x).

∴ [x−(2+√**3**)] [x−(2-√**3)**] = 0

(x−2−√**3**)(x−2+√**3**) = 0

On multiplying the above equation we get,

x^{2}-4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

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

Now, on further factorizing (x^{2}–2x−35) we get,

x^{2}**–(7−5)x −35** = x^{2}– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

**(x+5)(x−7) = 0**

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√**3** , 2-√**3**, **−5 and 7.**