Question

What must be subtracted from x cube-6x square-15x+80 so that the result is exactly divisible by x square+x-12

Answer 1

Refer to the above attachment !!

The required value that must be subtracted from the dividend is 4x-4 .

Answer 2

Heya!

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Let p(x) (divedend) = x^3 - 6x^2 - 15x + 80
g(x) (divisor) = x^2 + x - 12

Since dividend = divisor*quotient + remainder

=) p(x) = g(x) * q(x) + r(x)

=) p(x) - r(x) = g(x) * q(x)

Hence when remainder subtracted from p(x) , then it exactly divisible by g(x).

By division ;

=) x^2 +x-12 ) x^3 - 6x^2 - 15x + 80 ( x -7
x^3 + x^2 - 12x
- - +
___________________
× - 7x^2 - 3x + 80
- 7x^2 - 7x +84
+ + -
_____________________
× +4x - 4

Hence answer is 4x - 4.

Hope it helps uh!!