How do you find the remainder when $P(x) = x^4 - 9x^3 - 5x^2 - 3x+4$ is divided by x+3?

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Answer 1 · 2016-01-13T14:35:58

292

To divide by (x + 3 ) you don't have to divide by (x + 3 ) . Using the

Remainder Theorem you just have to evaluate P( - 3 ).

$P(- 3 ) = ( - 3 )^4 -9(- 3 )^3 - 5(- 3 )^2 - 3( - 3 ) + 4$

= 81 + 243 - 45 + 9 + 4

= 292

ie. remainder = 292

How do you find the remainder when P(x) = x^4 - 9x^3 - 5x^2 - 3x+4P(x) = x^4 - 9x^3 - 5x^2 - 3x+4 is divided by x+3?