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How to factor x^3+7x^2+15x+9=0

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This is of the form ax^3+bx^2+cx+d=0. You will need to find all the factors of d, and see which make the equation true, or equal to zero...

9=±1,3,9 (and also ±1/3, 1/9)

x=-3 makes the equation equal to zero so (x+3) is on factor.

Now you can divide the equation by (x+3) to find the other factors...

(x^3+7x^2+15x+9)/(x+3)
x^2 remainder 4x^2+15x+9
4x remainder 3x+9
3 remainder 0 so

(x+3)(x^2+4x+3) now you can factor the term in the parentheses...

(x+3)(x+1)(x+3)=0

(x+1)(x+3)^2=0 so x=-1, -3


User Jeff Martin
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