Given that the zeroes of the cubic polynomial \( x^{3}-6 x^{2}+3 x+10 \) are of the form \( a \), \( a+b, a+2 b \) for some real numbers \( a \) and \( b \), find the values of \( a \) and \( b \) as well as the zeroes of the given polynomial.

Given:

The zeroes of the cubic polynomial \( x^{3}-6 x^{2}+3 x+10 \) are of the form \( a \), \( a+b, a+2 b \) for some real numbers \( a \) and \( b \).

To find:

Here, we have to find the values of \( a \) and \( b \) as well as the zeroes of the given polynomial.

Solution:

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

$a,(a+b)$ and $(a+2 b)$ are the zeroes of $f(x)$.

This implies,

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

Therefore,

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

$3 a+3 b=6$

$a+b=2$.........(i)

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

This implies,

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

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

$2 a+2(2+b)+a(2+b)=3$

$2 a+2(2+2-a)+a(2+2-a)=3$

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

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

$a^{2}-4 a-5=0$

$a^2-5a+a-5 = 0$

$a (a - 5) + 1 (a - 5) = 0$

$(a - 5) (a + 1) = 0$

$a = 5$ or $a=-1$

This implies,

If $a = -1$, then $b = 2-(-1)=2+1=3$           [From (i)]

If $a = 5$, then $b=2-5=-3$                 [From (i)]

Therefore,

Required zeroes of $f(x)$ are, When, $a = -1, b = 3$, then

$a,(a+b),(a + 2) = -1, (-1+3), (-1+6)$

$=-1,2, 5$

When, $a = 5, b = -3$, then

$a, (a + b), (a + 2b) = 5, (5 -3), (5 -6)$

$=5,2,-1$

Hence, the required values of $a$ and $b$ are $a = - 1$ and $b= 3$ or $a = 5, b = -3$ and the zeroes are $-1,2$ and $5$.

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Updated on: 10-Oct-2022

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