document.write( "Question 376787: Factor to find all the real zeros of P(x)= 2x^3+5x^2-12x \n" ); document.write( "

Algebra.Com's Answer #268026 by jsmallt9(3758) $\"\"$ $\"About$

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$\"P%28x%29=+2x%5E3%2B5x%5E2-12x\"$
\n" ); document.write( "When factoring always start with the Greatest Common Factor (GCF), The GCF of F(x) is x. Factoring out x we get:
\n" ); document.write( " $\"P%28x%29=+x%282x%5E2%2B5x-12%29\"$
\n" ); document.write( "The second factor is a quadratic trinomial which factors as follows:
\n" ); document.write( " $\"P%28x%29=+x%282x-3%29%28x%2B4%29\"$

\n" ); document.write( "(Note: If you have trouble factoring trinomials like $\"2x%5E2%2B5x-12\"$ then use the Quadratic Formula. You will get -4 and 3/2 as answers from the Quadratic Formula. This makes (x- (-4)) (or (x+4)) and (2x - 3) factors. Note where the numerator and denominator of 3/2 ended up in (2x/3).)

\n" ); document.write( "P(x) is now fully factored. And from the factors we can find the zeros by setting each factor equal to zero and then solving for x:
\n" ); document.write( "x = 0 or 2x-3 = 0 or x+4 = 0
\n" ); document.write( "The first equation is already solved for x. Solving the other tow we get:
\n" ); document.write( "x = 0 or x = 3/2 or x = -4
\n" ); document.write( "These are the three zeros of P(x). \n" ); document.write( "

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