Marselamjmidn
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factor completely h^3+2h^2-9h-18
(please can someone explain to me the steps and not just the answer)

Factor Completely H32h29h18 Please Can Someone Explain To Me The Steps And Not Just The Answer class=

Sagot :

AlgebraicAardvarkidn

Step 1: Group this into two groups:

    [tex]h^3 + 2h^2-9h-18=(h^3 + 2h^2)+(-9h-18)[/tex]

When you  do that, be sure to group the "–" into the group with the 9 and make sure there's "+" between the groups.

Step 2: Factor out the GCF from both individual groups:

    [tex](h^3 + 2h^2)+(-9h-18) = h^2 (h+2) -9(h+2)[/tex]

Step 3: Notice both groups have that (h+2)?  That's a new GCF between the two groups, so factor that out:

    [tex]h^2 (h+2) -9(h+2) = (h+2)(h^2-9)[/tex]

Step 4: That [tex](h^2-9)[/tex] piece can be factored further, since it's a difference of squares:

    [tex](h+2)(h^2-9) = (h+2)(h+3)(h-3)[/tex]

Now it's factored.

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