Question

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

Answer 1

Answer:

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Answer 2

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→  The required values are

   a = 11  and  b = -6.

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Given : -

We are given to find the values of a and b if :

 $x^{2} - 3x + 2 \: \: divides \: \: x^{3} - 6x^{2} + ax + b$

 Solution : -

Factor theorem :

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

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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

$x^{2} - 3x + 2 = 0 \\\\(x-1) (x-2) = 0 \\\\x - 1 = 0 \: , \: x - 2 = 0 \\\\x = 1 , 2$

By factor theorem, we must get

$1^{2} - 6 \times 1^{2} + a \times 1 + b = 0 \\\\1 - 6 + a + b = 0 \\\\a + b = 5 \: \: \: \: ---- (i)$

and

$2^{3} - 6 \times 2^{2} + a \times 2 + b = 0 \\\\8 - 24 + 2a + b = 0 \\\\2a + b = 16 \: \: \: \: ------- (ii)$

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

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

And, from equation (i), we get

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

Thus, the required values are

   a = 11  and  b = -6