If the zeroes of the cubic polynomial

Solution:

Let $\quad f(x)=x^{3}-6 x^{2}+3 x+10$

Given that, $a_{1}(a+b)$ and $(a+2 b)$ are the zeroes of $f(x)$. Then,

Sum of the zeroes $=-\frac{\left(\text { Coefficient of } x^{2}\right)}{\left(\text { Coefficient of } x^{3}\right)}$

$\Rightarrow \quad a+(a+b)+(a+2 b)=-\frac{(-6)}{1}$

$\Rightarrow \quad 3 a+3 b=6$

$\Rightarrow \quad a+b=2 \quad \ldots$ (i)

Sum of product of two zeroes at a time $=\left(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}\right)$.

$\Rightarrow \quad a(a+b)+(a+b)(a+2 b)+a(a+2 b)=\frac{3}{1}$

$\Rightarrow a(a+b)+(a+b)\{(a+b)+b\}+a\{(a+b)+b\}=3$

$\Rightarrow \quad 2 a+2(2+b)+a(2+b)=3$  [using Eq. (i)]

$\Rightarrow \quad 2 a+2(2+2-a)+a(2+2-a)=3 \quad$ [using Eq. (i)]

$\Rightarrow \quad 2 a+8-2 a+4 a-a^{2}=3$

$\Rightarrow \quad-a^{2}+8=3-4 a$

$\Rightarrow \quad a^{2}-4 a-5=0$

Using factorisation method,

a2-5a+a-5 = 0 =3

⇒ a (a – 5) + 1 (a – 5) = 0

⇒ (a – 5) (a + 1) = 0

⇒ a = -1, 5

when a = -1, then b = 3

When a = 5, then b = – 3                                                                                      [using Eq. (i)]

∴Required zeroes of f(x) are

When       a = -1 and b = 3

then,       a,(a+b),(a + 2) = -1, (-1+3), (-1+6) or -1,2, 5

When      a = 5and b = -3 then

a, (a + b), (a + 2b) = 5, (5 -3), (5 -6) or 5,2,-1.

Hence, the required values of a and b are a = – 1 and d = 3ora = 5, b = -3 and the zeroes are -1,2 and 5.