Question

If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’

Answer 1

find the roots of
${x}^{2} - 3x + 2$
then place these values in the second polynomial,you will get two equation,solve them for calculating a and b.
${x}^{2} - 3x + 2 = 0 \\ {x}^{2} - 2x - x + 2 = 0 \\ x(x - 2) - 1(x - 2) = 0 \\ (x - 2)(x - 1) = 0 \\ x - 2 = 0 \\ x = 2 \\ x - 1 = 0 \\ x = 1$
put x = 2
${x}^{3} - 6 {x}^{2} + ax + b = 0 \\ 8 - 6(4) + 2a + b = 0 \\ 8 - 24 + 2a + b = 0 \\ 2a + b = 16$
put x= 1
${x}^{3} - 6 {x}^{2} + ax + b = 0 \\ 1 - 6 + a + b = 0 \\ a + b = 5$
subtract both equations
$2a - a \: = 16 - 5 \\ a = 11$
$a + b = 5 \\ 11 + b = 5 \\ b = 5 - 11 \\ b = - 6$

Answer 2

Answer:  The required values are

a = 11  and  b = -6.

Step-by-step explanation:  We are given to find the values of a and b if :

$x^2-3x+2~~~\textup{divides}~~~x^3-6x^2+ax+b.$

Factor theorem :  If a factor (x - a) divides a polynomial f(x), then f(a) = 0.

First, we will factorize the expression $x^2-3x+2$ as follows :

$x^2-3x+2\\\\=x^2-2x-x+2\\\\=x(x-2)-1(x-2)\\\\=(x-1)(x-2).$

So, we get

$x^2-3x+2=0\\\\\Rightarrow (x-1)(x-2)=0\\\\\Rightarrow x-1=0,~~x-2=0\\\\\Rightarrow x=1,2.$

By factor theorem, we must get

$1^3-6\times 1^2+a\times1+b=0\\\\\Rightarrow 1-6+a+b=0\\\\\Rightarrow a+b=5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$2^3-6\times2^2+a\times2+b=0\\\\\Rightarrow 8-24+2a+b=0\\\\\Rightarrow 2a+b=16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Subtracting equation (i) from equation (ii), we get

$(2a+b)-(a+b)=16-5\\\\\Rightarrow a=11.$

And, from equation (i), we get

$11+b=5\\\\\Rightarrow b=5-11\\\\\Rightarrow b=-6.$

Thus, the required values are

a = 11  and  b = -6.