Question

if x³-6x²+ax+b is exactly divisible by x²-3x+2, then the value of a is​

Answer 1

Step-by-step explanation:

Finds the roots of

${x}^{2} - 3x + 2 = 0$

then place these values in the second polynomial,you will get two equation,solve them for calculating a and b.

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

Put x=2

$= \geqslant {x}^{3} - 6 {x}^{2} + ax + b = 0 \\ 8 - 6(4) + 2 + b = 0 \\ 8 - 24 + 2a + b = 0 \\ 2a + b = 16$

$put \: x = \geqslant 1$

$= \geqslant {x}^{3} - 6 {x}^{2} + a + b = 0 \\ 1 - 6 + a + b = 0 \\ a + b = 5$

$subtract \: both \: equations........$

$= \geqslant 2a - a = 16 - 5 \\ a = 11 \\ a + b = 5 \\ 11 + b = 5 \\ b = 5 - 11 \\ b = - 6$

$hope \: it \: helps.....$