SOLUTION: if the function x^3 + 2x^2 - ax - 8 is evenly divisible by (x+2), find the value of a

Question 1192990: if the function x^3 + 2x^2 - ax - 8 is evenly divisible by (x+2), find the value of a
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52810) (Show Source): You can put this solution on YOUR website!
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if the function x^3 + 2x^2 - ax - 8 is evenly divisible by (x+2), find the value of a
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According to the Remainder theorem, the number -2 is the root of the polynomial. So, substitute this value of -2 into the polynomial. It will give you an equation for the unknown coefficient "a" (-2)^3 + 2*(-2)^2 - a*(-2) - 8 = 0, or -8 + 2*4 + 2a - 8 = 0, 2a = 8 a = 8/2 = 4. ANSWER. a = 4.

Solved.

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!

x^2 + 0x - a x + 2)x^3 + 2x^2 - ax - 8 x^3 + 2x^2 0x^2 - ax 0x^2 + 0x -ax - 8 -ax -2a -8+2a The remainder should equal 0, so -8+2a = 0 2a = 8 a = 4 Edwin

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