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I need to factor these problems but need it explain in words on how to do it so I can...

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gsstom | Student, College Freshman | (Level 2) Honors

Posted September 7, 2013 at 8:24 PM via web

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I need to factor these problems but need it explain in words on how to do it so I can work on other ones. Please help by not just solving but step by step instructions.

9w-w^3

2a^2+3a+1

Thanks.

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baxthum8 | High School Teacher | (Level 3) Associate Educator

Posted September 7, 2013 at 8:41 PM (Answer #1)

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In order to factor an algebraic expression, the first step is to always factor (divide out) the greatest common factor if the terms have one. For instance in the first example listed of:

`9w - w^3` both terms have a common factor of w.  Therfore we will pull out a w from each term.  This will give us:

`w (9 - w^2)` Notice that if I was to distribute the w back through I would have the original expression because w * 9 = 9w and w * w^2 = w^3.

Now in the expression in which I factored the w I'm left with what is called the difference of 2 squares inside the parenthesis.  This is because they (the 9 and w^2) are both perfect squares.

Difference of 2 squares rule is:  `(a^2 - b^2) = (a + b)(a - b)`

Therefore, in continuing to factor the parenthesis and keeping the w I had already factored, I will get:

`w (3 - w) (3 + w).`

In the second example:  `2a^2 + 3a + 1,`

there is no greatest common factor amongst all the terms.  So next I will factor the trinomial.  If factorable all trinomials will factor to 2 binomials.

Example:  `(a + b) (a + c) = a*a + a*c + b*a + b*c`

So, we need to terms that when multiplied together gives us `2a^2`

This can only be `2a* a`

`(2a + x) (a + y)` Now we need to determine what x and y will be.

However, when we distribute outer and inner terms, they must add to equal 3a.  However, also we need to find 2 integer values that when multiplied gives us the last term.  Since this is 1, the only possibilities are 1 and 1.

So, when factored, we get:  `(2a + 1) (a + 1).`

Notice that the product of the outer terms (2a*1=2a) and the product of the inner terms (1 * a = 1a) together have a sum of 3a, which is our middle term in ourr trinomial that we factored.

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