Question 92844: I really need help on this problem. I haven't taken algebra in years so im a little rusty.
Q: What is the remainder when x^3+5x^2-6x+10 is divided by x+3?
The book says that the answer is 46 but i'm not sure how to get that.
Thanks alot.
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! remember long division? ... you can use the same technique with polynomials
372 is the same as 3(10^2)+7(10)+2 ... just like 3x^2+7x+2
so, dividing x+3 into x^3+5x^2-6x+10 ... x goes into x^3, x^2 times ... (x^2)(x+3) is x^3+3x^2
(x^3+5x^2-6x+10)-(x^3+3x^2) is 2x^2-6x+10 ... x goes into 2x^2, 2x times ... (2x)(x+3) is 2x^2+6x
(2x^2-6x+10)-(2x^2+6x) is -12x+10 ... x goes into 12x+10, -12 times ... (-12)(x+3) is -12x-36
(12x+10)-(-12x-36) is 46 ... this is the remainder (it is not divisible by x)
this is done like long division ... with all the same power terms lined up vertically
not the case here, but if you were missing a term; like x^3+x-7 (no x^2 term); you would put in a zero for a place holder x^3+0x^2+x-7 ... to keep things properly aligned
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