Factoring gets really interesting when we start to deal with more complicated algebraic polynomials, which are just big chains of algebraic expressions. One of these is called a binomial (you might have encountered them already), which is any polynomial with just two terms, usually not alike. The 'bi' in the word gave it away didn't it! For example $x+y$x+y, $4-w$4−w, $-z+55$−z+55 are all binomials.
Really interesting patterns appear when we multiply two binomials together, and you here we're going to look at one of the form $\left(a+b\right)\left(a-b\right)$(a+b)(a−b), which you might remember as being called difference of two squares.
So now we know that:
$\left(a+b\right)\left(a-b\right)=a^2-b^2$(a+b)(a−b)=a2−b2
Be careful that in this example where $a$a is the same sign in both brackets the answer is always $a^2-b^2$a2−b2 and NOT $b^2-a^2$b2−a2! We already know how to use the above property to expand the bracketed expression on the left, but now we can also use it to factor the one on the right, since factoring is the opposite of expansion.
Answer the following.
Expand $\left(x+5\right)\left(x-5\right)$(x+5)(x−5).
Hence factor $x^2-25$x2−25.
Factor $121m^2-64$121m2−64.
Factor $3t^2-12$3t2−12.
Factor the expression $16x^2-81y^2$16x2−81y2.
Think about how we should go from $a^2$a2 and $b^2$b2 terms in the property above to $a$a's and $b$b's by finding square roots.
Do
$\sqrt{16x^2}$√16x2 | $=$= | $\sqrt{16}\sqrt{x^2}$√16√x2 |
$=$= | $4x$4x | |
$\sqrt{81y^2}$√81y2 | $=$= | $\sqrt{81}\sqrt{y^2}$√81√y2 |
$=$= | $9y$9y | |
Therefore | ||
$\left(4x\right)^2$(4x)2 | $=$= | $16x^2$16x2 |
and | ||
$\left(9y\right)^2$(9y)2 | $=$= | $81y^2$81y2 |
So | ||
$16x^2-81y^2$16x2−81y2 | $=$= | $\left(4x+9y\right)\left(4x-9y\right)$(4x+9y)(4x−9y) |
Factor the difference of squares of the form x^2 – a^2 (e.g., x^2 – 16)