Factoring Quadratics
In the previous lesson, we learned how to factor polynomials using the greatest common factor method and the factor by grouping method to undo distributing and double-distributing/FOILing. There's one type of problem that we didn't talk about though! In some double-distributing problems, the result has like terms that get combined. For example, if you multiply out $(2x-5)(x-9)$, you get $2x^2-5x-18x+45$, but the two middle terms combine to give us $2x^2-23x+45$ as a final answer. But if we try to go the other direction, we run into problems - factoring by grouping doesn't seem to work on $2x^2-23x+45$, because the terms can't be split up into two equal groups!
In order to use factoring by grouping on an expression like this one, we have to split up one of the terms into two pieces first. The problem is, we have to find exactly the right way to split it up - if we do it wrong, then the two groups won't end up with the "matching leftovers" that are the key to factoring by grouping. In the case of $2x^2-23x+45$, the -23x has to be split up into -5x and -18x specifically; if we tried to split it up using another combination, it won't work! For example, suppose we tried using -20x and -3x instead. If we do that, then $2x^2-20x$ becomes $2x(x-10)$, and $-3x+45$ factors to $-3x(x-15)$. Since $(x-10)$ and $(x-15)$ don't match, we can't pull them out as a common factor, so we're stuck!
So, if there's only one specific combination that will work to split up the expression, how can we find what that combination is? In the example above, we knew to turn -23x into -5x and -18x because we had done the problem in the opposite direction first - but what if we're doing a problem where we don't already have that information, like $-8x +4x^2 + 3 $? The good news is, even though we don't know exactly what numbers were used to get this expression, we do have enough information to find out. Since all the terms in the expression came from the same sources in different combinations, we can use the other terms to help us determine how to split it up.
The first thing we need to do is put the expression into standard form - in other words, rearrange it so that the terms are in order from the highest power of x to the lowest. In the case of $-8x +4x^2 +3$, the highest power is $x^2$, then $x$, then constant (no x's), so we rewrite it as $4x^2 -8x +3$. Once it's in standard form, the middle term is going to be the one that's getting split up, and we'll use the other two terms to help us figure out how to split it.
To split up that middle term, we need to find a pair of numbers that adds up to the coefficient (the number in front of x). In this case, that means we need ___ + ___ = -8. There are so many different combinations that could work for that: -4 and -4, -18 and +10, -7 and -1, -3 and -5, 1 and -9, -1000 and +992...
However, to be the correct pair, the two numbers also need to multiply out to the same thing that the other two terms' coefficients multiply do—for this problem, that means they have to multiply to the same thing as 4 × 3. In other words, __ × ___ = 12.
The nice thing is, there are a lot fewer combinations that will work for that multiplication requirement! Assuming that the two numbers can only be integers and that it doesn't matter which order they go in, the only options for ___×___=12 are 1×12, 2×6, 3×4, -1×-12, -2×-6, or -3×-4.
Now we just have to find one of these pairs that also meets the addition requirement of adding up to -8. It turns out that one set does: -2 and -6. Therefore, if we split -8x into -2x and -6x, we should end up with something that will work when we factor by grouping.
Let's test it! $4x^2 -8x+3$ becomes $(4x^2 -2x)+ (-6x+3)$, then the first group becomes $2x(2x-1)$ and the second group becomes $-3(2x-1)$. The $(2x-1)$ matches on both—it worked! So we factor it out to give us $(2x-1)(2x-3)$ as the final result.
To sum up: if you have an expression in the form $ax² + bx +c$, you may be able to factor it by grouping if you split up the bx term, using a pair of numbers that add up to b and multiply to the same thing as a×c. To do that, list out pairs of numbers that multiply to a×c, then find which of those pairs add up to b. Use that pair to rewrite the expression as $ax^2+\_x+\_x+c$, then use factoring by grouping. Factor out the GCF of the first two terms and the GCF of the last two terms, and the "leftovers" in parentheses should be the same for both groups. Factor out the "leftovers," adding the two GCFs together in a second set of parentheses, and you're done!
Unfortunately, not every quadratic expression will be factorable. If you've checked all the factor pairs that multiply to a×c, and none of them add up to b, then that means that particular equation can't be factored. For example $3x^2 +90x-14$ can't be factored, because none of the factor pairs of $3×-14$ (or $-42$) add up to 90.
Let's do one more example: factoring $24x²+46x+7$.
First, we need to find a pair of numbers that multiply to 24×7 (which is 168), and add up to 46. So we look at all the factor pairs of 168, and we find that the pair that works is 4 and 42.
Next, we use that pair to split up the middle term, rewriting the expression as 24x²+4x+42x+7. Now we factor by grouping: we group them into pairs as (24x²+4x)+(42x+7), then factor out 4x from the first group and 7 from the second group, giving us 4x(6x+1)+7(6x+1). Finally, we pull out (6x+1) and combine the 4x and +7 into parentheses, resulting in (4x+7)(6x+1) as the final answer.
Now it's your turn! Try these problems on your own and let me know your answers in the comments.
Factor $2x²+5x-3$
Factor $8m²+2m-21$
Factor $16x²-25$ (Hint: since there's no x term, that means b=0!)
