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Basics of factoring
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The word “factor” means “multiplier”. When used as a verb, “factor&rdsquo; means “to express as a product”. We should always proceed in this order (from easiest to most difficult) and repeat this time after time until the expression is fully factored.
Taking a common factor
A common factor is a multiplier that is common to two or more terms. Taking a common factor typically means taking the greatest common divisor of all terms. Consider the following example.
\begin{align} 30a^2b-5ab &= (5ab)(6a)-(5ab)(1) \tag{You may skip this step} \\ &=5ab(6a-a) \end{align}What remains in the brackets can be obtained by dividing each term by the common factor. In other words, we can do the factoring as follows.
\begin{align} 30a^2b-5ab &= \left(\dfrac{30a^2b}{5ab}-\dfrac{5ab}{5ab}\right) \tag{You may skip this step} \\ &=5ab(6a-a) \end{align}Factoring by grouping
With this method, we divide the expression into groups of terms, and we take a common factor from each of the groups. The target is to end up with the same set of brackets, then we can take this as a common factor in a subsequent step. Consider the expression \(3a^2+3ab-5a-5b\).
\begin{align} \boxed{3a^2+3ab}-\boxed{5a-5b} &= 3a\underline{(a+b)}-5\underline{(a+b)} \\ &=(a+b)(3a-5) \end{align}Special quadratics
Difference of squares
This is always (as the name tells) ‘a square minus a square’, so it is always made of two terms. The general pattern is
Consider the following as an example.
\begin{align} 4x^2-9y^2&=(2x)^2-(3y)^2 \tag{You may skip this step} \\ &=(2x-3y)(2x+3y) \end{align}Perfect square
This is the expansion of a binomial squared. The expansion is \((a\pm b)^2=a^2\pm 2ab+b^2\). We can go from right to left as much as from left to right. So, the general pattern is
The expression is always made of three terms. The first and last are always positive (because they are squares). The second can either be positive or negative; hence the \(\pm) (pronounced ‘plus or minus’). Note that the sign of the second term matches the sign inside the squared bracket.
We have to check whether a trinomial is a perfect square. This is how we do it.
- The first and last terms are positive and are perfect squares
- The second term is double the product of the square roots of the first and third terms.
Consider the following example.
\begin{align} 9x^2-24xy+16y^2 &= (3x)^2-(2)(3x)(4y)+(4y)^2 \tag{You may skip this step} \\ &= (3x-4y)^2 \end{align}The expression \(x^2-2x-1\) is not a perfect square because the last term is negative. The expression \(x^2-3x+1\) is not a perfect square because the middle term is not double \(\sqrt{x^2}\) and \(\sqrt{1}\).
Regular quadratics
A quadratic in standard form \(ax^2+bx+c\) can be factored by trying to find two integers that
- multiply to \(ac\)
- add up to \(b\)
If \(c=0\), the expression can be factored simply by taking \(x\) as a common factor from the first two terms. If the two integers are not easy to find, list all pairs of divisors of \(ac\). Consider as an example \(6x^2+7x-20\). The product \(ac=(6)(-20)=-120\). To find possible integers, we list all pairs of divisors of \(-120\) systematically so we do not miss anything. Once we find ourselves repeating the divisors but in the reverse order, we should stop listing them, because we will only get another copy of what we already found.
Because the product \(-120\) is negative, one of the divisors must be positive and the other negative. This is what we mean by \(\pm\) in the first column and \(\mp\) (pronounced ‘minus or plus’) in the second column.
We find that \(+15\) and \(-8\) are the two integers that add up to \(7\). Therefore, we split the middle term according to these two numbers, because \(7x=15x-8x\). Then we factor by grouping.
\begin{align} 6x^2+7x-20 &= 6x^2+15x-8x-20 \\ &= 3x(2x+5)-4(2x+5) \\ &= (2x+5)(3x-4) \end{align}\(\pm 1\) | \(\mp 120\) |
\(\pm 2\) | \(\mp 60\) |
\(\pm 3\) | \(\mp 40\) |
\(\pm 4\) | \(\mp 30\) |
\(\pm 5\) | \(\mp 24\) |
\(\pm 6\) | \(\mp 20\) |
\(\pm 8\) | \(\mp 15\) |
\(\pm 10\) | \(\mp 12\) |
\(\pm 12\) | \(\mp 10\) |
\(\vdots\) | \(\vdots\) |