Section 4.6 Factoring Binomials
¶Difference of Squares.
A binomial (two term polynomial) of form a^2-b^2 always factors into the product (a+b)(a-b)\text{.} We can confirm this by applying FOIL to the expression (a+b)(a-b)\text{.}
A few simple examples follow. As always, we can check our result by expanding the factored expression.
Now let's consider a few expressions that don't immediately fit the pattern. Consider x^{10}-16\text{.} Hopefully we are quick to see that 16 is the square of 4\text{.} To use our factor pattern successfully, we need to also recognize that x^{10} is a perfect square, as is any even power of x\text{.} The power-to-a-power rule of exponents relates that (x^m)^n=x^{mn}\text{.} So the power of x we square that results in x^{10} must be half of 10\text{,} i.e. 5\text{.} Putting it all together we have:
Similar examples follow.
Exploration 4.6.1.
Exploration 4.6.2.
Sum and Difference of Cubes.
A binomial (two term polynomial) of form a^3-b^3 always factors into the product (a-b)(a^2+ab+b^2)\text{.} We can confirm this by expanding the expression (a-b)(a^2+ab+b^2)\text{.}
Similarly, a binomial of form a^3+b^3 always factors into the product (a+b)(a^2-ab+b^2)\text{.} We can confirm this by expanding the expression (a+b)(a^2-ab+b^2)\text{.}
Example 4.6.3.
Factor 8x^3+27 and 8x^3-27\text{.}
For both binomials, \(8x^3\) corresponds to what is identified in the patterns as \(a^3\) and \(27\) corresponds to what is identified in the pattern as \(b\text{.}\) The resultant expressions for \(a\text{,}\) \(b\text{,}\) \(a^2\text{,}\) and \(b^2\) and shown in Figure 4.6.4 and the factorizations are shown to the left of the table.
Example 4.6.5.
Factor the binomials 1+64x^{15} and 1-64x^{15}\text{.}
Note that the power-to-a-power rule of exponents gives us \((x^5)^3=x^{(5\times 3)}\text{.}\)
| \(a^3=1\) | \(b^3=64x^{15}\) |
| \(a=1\) | \(b=4x^5\) |
| \(a^2=1\) | \(b^2=16x^{10}\) |
| \(ab=4x^5\) |
Exploration 4.6.3.
Exploration 4.6.4.
Sum of Squares.
Unless the expression also happens to be a sum of cubes, sums of squares do not factor - that is, they are prime.
x^2+4 is prime.
y^4+25 is prime.
w^6+4x^2 is prime.
Many folks would like x^2+4 to factor, so much so that they will write x^2+4=(x+2)^2\text{.} Would that it were so. But alas:
In summary, x^2+4\neq (x+2)^2\text{,} x^2+4x+4=(x+2)^2\text{.}
Exercises Exercises
Factor each binomial after first completing the indicated table.
1.
Use the factor pattern a^2-b^2=(a+b)(a-b) to factor x^{10}-25y^4 after first completing the entries in Figure 4.6.9
2.
Use the factor pattern a^3-b^3=(a-b)(a^2+ab+b^2) to factor 8x^3-y^6 after first completing the entries in Figure 4.6.11
3.
Use the factor pattern a^3+b^3=(a+b)(a^2-ab+b^2) to factor 125t^{12}+27x^9 after first completing the entries in Figure 4.6.13
Factor each binomial. Check your result by expanding the factored expression. If the binomial does not factor, state that it is prime.