Above we have seen that (2x2 - x + 3) and (3x3 + x2 - 2x - 5) are the factors of 6x5 - x4 + 4x3 - 5x2 - x - 15. In this case we are able to find one factor given the other one. How are we going to solve in case when we are not given either of them. Finding the factors of a given expression forms the part of our attention now. First, we look at binomial expressions and once we understand this we move on to trinomials and polynomials. If the given expression is in the form of an identity (we look at them shortly) our job becomes easier otherwise we have to adopt trial and error method until we get at least one of the factors. Once we know one of the factors then by employing the division method we can get other factors.
Example
Factorize x2 + 6x + 9.
If we substitute x = 1, the value of the expression will be (1)2 + 6(1) + 9 = 16
Since the value of the numerical expression is not 0, we substitute another value. We will continue to do so until we get a zero.
If we substitute x = -1, the value of the expression will be (-1)2 + 6(-1) + 9 = 4
If we substitute x = 2, the value of the expression will be (2)2 + 6(2) + 9 = 25
If we substitute x = -2, the value of the expression will be (-2)2 + 6(-2) + 9 = 1
If we substitute x = 3, the value of the expression will be (3)2 + 6(3) + 9 = 36
We substitute x = -3, the value of the expression will be (-3)2 + 6(-3) + 9 = 0
For x = -3, the value of the expression is 0. That is, x + 3 is one of the factors of the expression x2 + 6x + 9. To obtain the other factor we divide the expression by the factor we obtained. That will be
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x + 3 )
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x2 + 6x + 9
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( x + 3
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(-)
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x2 + 3x
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3x + 9
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(-) 3x + 9
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0
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From the division, we observe that x + 3 is the other factor. When this is equated to zero we obtain x = - 3. Therefore, the factors of x2 + 6x + 9 are (x + 3)(x + 3) or (x + 3)2.
In the above example we note that x2 + 6x + 9 = (x + 3)2. Isn't this identical to a2 + 2ab + b2 = (a + b)2? The value of 'a' being x and that of 'b' equal to 3. This is one of the basic identities we get to see in algebra.