Question

what must be subtracted from x^3-6x^2-15x+80 so that the result is exactly divisible by x^2+x-12 by factor theorem​

Answer 1

Answer:

So, from the condition given in the question, we get to know that the dividend should be ${{x}^{3}}-6{{x}^{2}}-15x+80$ and the divisor should be ${{x}^{2}}+x-12$. ... Hence, (4x - 4) is the term that must be subtracted from ${{x}^{3}}-6{{x}^{2}}-15x+80$ so that the result is exactly divisible by ${{x}^{2}}+x-12$.

Step-by-step explanation: