# 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.

embizze | High School Teacher | (Level 1) Educator Emeritus

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## 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)`