Solve each polynomial equation by factoring.
1.
5x³+10x²+5x = 0
Factor out 5x
5x(x²+2x+1) = 0
Factor the quadratic in the parentheses:
5x(x+1)(x+1) = 0
5x=0; x+1=0; x+1=0
x=0; x=-1 x=-1
For #1 - Would this answer be cannot be factored ?
No, for we factored it twice.
Because I did it, and i'm getting x = {0, -1, -1}.
That's the correct solution but the only way to get it
is to factor the polynomial.
2.
x³+2x²-9x-18 = 0
Factor x² out of the first two terms on the left
and factor -9 out of the last two terms on the left:
x²(x+2)-9(x+2) = 0
We factor out the common factor (x+2)
(x+2)(x²-9) = 0
Factor the expression in the second parentheses as
the difference of two squares:
(x+2)(x-3)(x+3) = 0
Use the zero-factor property:
x+2=0; x-3=0; x+3=0
x=-2; x=3; x=-3
Edwin
