Posted by jezebel on Friday, September 28, 2012 at 2:21pm.
I think what you're being asked is this. There's some polynomial - let's call it f(x) - that when divided by (x + 10) equals (x^2 - 6x + 10) with remainder -1. What is f(x)?
If I'm right, then we should be able to find it by multiplying (x^2 - 6x + 10) by (x + 10), and then adding -1 to the result. Let's do that:
To multiply (x^2 - 6x + 10) by (x + 10), multiply (x^2 - 6x + 10) by x, then multiply (x^2 - 6x + 10) by 10, and add the two together. That is:
The product of (x^2 - 6x + 10) and x is (x^3 - 6x^2 + 10x).
The product of (x^2 - 6x + 10) and 10 is (10x^2 - 60x + 100).
The sum of the above two expressions is (x^3 + 4x^2 - 50x + 100). Now add that -1 to it to allow for the remainder you were told about earlier: that gives (x^3 + 4x^2 - 50x + 99).
I think that's the answer, so now let's check it. If I'm right, then (x^3 + 4x^2 - 50x + 99) divided by (x + 10) should equal (x^2 - 6x + 10) with remainder -1. Does it?
To divide (x + 10) into (x^3 + 4x^2 - 50x + 99), ask what is x^3 divided by x? The answer is x^2, so that's the first (squared) term of my quotient. Next, multiply (x + 10) by x^2 and subtract the result from (x^3 + 4x^2 - 50x + 99), just as you would when doing the first step of a long division sum:
(x^3 + 4x^2 - 50x + 99) minus (x^3 + 10x^2) equals (-6x^2 - 50x + 99).
To divide (-6x^2 - 50x + 99) by (x + 10), ask what is -6x^2 divided by x? The answer is -6x, which is the second (linear) term of my quotient. Next, multiply (x + 10) by -6x and subtract the result from (-6x^2 - 50x + 99), again just as you would when doing the second step of a long division sum:
(-6x^2 - 50x + 99) minus (-6x^2 - 60x) equals (10x + 99).
Finally, (10x + 99) is 10 times (x + 10) remainder -1. So 10 is the third (constant) term of my quotient. So the complete answer is the first (squared) term of my quotient plus the second (linear) term of my quotient plus the third (constant) term of my quotient, which is (x^2 - 6x + 10) remainder -1. So I think I've got it right.
Does that help? It would be easier to express the above if I could write it out like a proper long division sum, but I'm hoping you can relate the above to the examples you'll have seen in your class, which will probably have been written out this way.