Question

the expression ax^3-8x^2+bx+6 is divisible by x^2-2x-3.Find the values of a and b.

Answer 1

@ShailKumar

Answer 2

Can you try to divide ?

Answer 3

hw.. there are unknowns

Answer 4

if it is divisible by x^2-2x-3,
then it will be divisible by its individual factors : x-3 and x+1 also since x^2-2x-3 = (x-3)(x+1)

Answer 5

by remainder/factor theorem,
x-3 is a factor => f(3) = 0
x+1 is a factor => f(-1) = 0

use them to solve a, b

Answer 6

I need to form two equations?

Answer 7

\(ax^3-8x^2+bx+6\)

\( = (mx + n)(x^2-2x-3)\)

\(= mx^3 - 2 mx^2 - 3mx + nx^2 - 2nx - 3n \)

\(= mx^3 + (-2m + n)x^2 + (-3m - 2n)x - 3n \)

-3n = 6
n = -2

-3m - 2n = b
-3m - 2(-2) = b
-3m + 4 = b Eq. 1

-2m + n = - 8
-2m + (-2) = -8
-2m = -6
m = 3

Eq. 1: -3m + 4 = b
-3(3) + 4 = b
-9 + 4 = b
b = -5

a = m
a = 3

Answer 8

A colorful explanation of the lines above.

Line 1: \(\color{red}{a}x^3+ \color{blue}{-8}x^2+\color{green}{b}x+\color{brown}{6}\)
Line 4: \(= \color{red}{m}x^3 + \color{blue}{(-2m + n)}x^2 + \color{green}{(-3m - 2n)}x + \color{brown}{- 3n}\)

Answer 9

Hw you form the second line..
(mx+n)(x^2-2x-3)
@mathstudent55

Answer 10

Since you are told a third degree polynomial is divisible by a second degree polynomial, then the quotient must be a first degree polynomial, which we know is a binomial of the form mx + n.

Answer 11

Ok..

Answer 12

Thanks @mathstudent55 @ganeshie8 @ShailKumar

Answer 13

You're welcome.