Question

what must be subtracted fromx3-6x2-15x+80 so that it is divisible by x²+x-12?

Answer 1

Divide is in the attachment
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We know that

Dividend= quotient × remainder

Dividend - remainder = quotient × divisor

Dividing x³-6x²-15x+80 by x²+x-12 we get

Quotient = x-7
Remainder = 4x-4
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If you subtract the remainder 4 x - 4 from x³-6x²-15x+80 , it will be divisible by x²+ x-12
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Hope this will help you....

Answer 2

Solution :-

x³ - 6x² - 15x + 80 ÷ x² + x - 12

The solution is attached regarding the answer of this question. Please have a look at it.

First we will divide x³ - 6x² - 15x + 80 by x² + x -12. After dividing, we will get 4x - 4 as the remainder. 

Then we will subtract 4x - 4 from x³ - 6x² - 15x + 80.

Then we will get x³ - 6x² - 19x + 84, which is exactly divisible by x² + x - 12 giving x - 7 as quotient.