# If ax³+3x²+bx-3 has a factor (2x-3) and leaves a remainder -3 when divided by (x+2), find the values of a and b and factorize the given expression.

*print*Print*list*Cite

### 1 Answer

## If ax³+3x²+bx-3 has a factor (2x-3) and leaves a remainder -3 when divided by (x+2), find the values of a and b and factorize the given expression.

Since 2x-3 is a factor, then x=3/2 is a root. We use synthetic division with 3/2 :

3/2 | a 3 b -3

|---------------------------------------------------------

a 3/2a+3 9/4a+9/2+b 27/8a+27/4+3/2b-3

Now since 3/2 is a root we have 27/8a+3/2b+15/4=0 or 9a+4b=-10.

We can also use synthetic division with -2 as a divisor:

-2 | a 3 b -3

|---------------------------------------------------

a 3-2a -6+4a+b 12-8a-2b-3

Since x+2 leaves a remainder of -3 we have -8a-2b+9=-3

or 4a+b=6.

We now solve 9a+4b=-10 and 4a+b=6 simultaneously to get a=34/7 and b=-94/7.

Then `ax^3+3x^2+bx-3` becomes `34/7 x^3+3x^2-94/7 x -3` . Factoring out a 1/7 yields `1/7(34x^3+21x^2-94x-21)` . Since 2x-3 is a factor we can use synthetic division or polynomial long division to get:

`1/7(2x-3)(17x^2+36x+7)` , which does not factor further in the reals.

**Then the answer is a=34/7, b=-94/7, and the factored form is ****`1/7(2x-3)(17x^2+36x+7)` **