polynomials qn (1 Viewer)

Masaken

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yeah oops, i think i had a mindblank there (looking back i think i was thinking about the remainder theorem even though the remainder theorem is only for linear factors 💀 💀 💀 ). thank you!!
 

Sstormwolf22

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I think you can still use remainder theorem?

Factorise the divisor into

Using remainder theorem in the original polynomial, we know that:
and

Using the polynomial in division form:


We know that
, so sub it into the divison form to get:



( does not work since which does not enable you to solve for a)
 

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