Recall: Factoring is unmultiplying.
We're going to unmultiply expressions like
(A) $x^2+3x+2$
(B) $x^2+5x+6$
(C) $x^2+4x+4$
Pop Quiz: Which factorization goes with each polynomial above? ( Hint: multiply out each option.)
(1) $(x+2)(x+3)$
(2) $(x+1)(x+2)$
(3) $(x+2)(x+2)$
(A) $x^2+3x+2=(x+1)(x+2)$ (2)
(B) $x^2+5x+6=(x+2)(x+3)$ (1)
(C) $x^2+4x+4=(x+2)(x+2)$ (3)
(B) $x^2+5x+6=(x+2)(x+3)$ (1)
(C) $x^2+4x+4=(x+2)(x+2)$ (3)
A Method for Factoring $x^2+bx+c$
We want to change $x^2+bx+c$ into an expression of the form $$(x+r)(x+s).$$
Multiplying this expression out we get $x^2+(r+s)x+rs.$
The Big Deal: To factor $x^2+bx+c$, all you need to do is find two numbers whose sum is $b$ and whose product is $c$.
Examples: Factor the following trinomials.
$t^2+11t+30$
We want two numbers that add to add to $11$ and multiply to be $30.$
$$
(t+\mbox{___})(t+\mbox{___})
$$
and fill in the blanks with these numbers.
Let's find numbers which do this by constructing a table of factorizations of $30.$ $$ \begin{array}{ll|l} \displaystyle &\displaystyle\mbox{Factors} &\mbox{Sum}\\\hline \displaystyle 30 &\displaystyle= 1\cdot 30 & 31\mbox{}\\ \displaystyle &\displaystyle= 2\cdot 15 & 17\mbox{}\\ \displaystyle &\displaystyle= 3\cdot 10 & 13\mbox{}\\ \displaystyle &\displaystyle= 5\cdot 6 & 11\, \checkmark \mbox{}\\ \end{array} $$ Since $5$ and $6$ add to $11$ and multiply to be $30,$ we have our factorization. $$t^2+11t+30=(t+5)(t+6)$$ Check
$$ \begin{array}{lll} \displaystyle (t+5)(t+6)&\displaystyle=t^2+6t+5t+30 &\mbox{using FOIL shortcut}\\ \displaystyle &\displaystyle= t^2+11t+30& \checkmark \mbox{}\\ \end{array} $$
Let's find numbers which do this by constructing a table of factorizations of $30.$ $$ \begin{array}{ll|l} \displaystyle &\displaystyle\mbox{Factors} &\mbox{Sum}\\\hline \displaystyle 30 &\displaystyle= 1\cdot 30 & 31\mbox{}\\ \displaystyle &\displaystyle= 2\cdot 15 & 17\mbox{}\\ \displaystyle &\displaystyle= 3\cdot 10 & 13\mbox{}\\ \displaystyle &\displaystyle= 5\cdot 6 & 11\, \checkmark \mbox{}\\ \end{array} $$ Since $5$ and $6$ add to $11$ and multiply to be $30,$ we have our factorization. $$t^2+11t+30=(t+5)(t+6)$$ Check
$$ \begin{array}{lll} \displaystyle (t+5)(t+6)&\displaystyle=t^2+6t+5t+30 &\mbox{using FOIL shortcut}\\ \displaystyle &\displaystyle= t^2+11t+30& \checkmark \mbox{}\\ \end{array} $$
We'll find numbers which add to $-13$ and multiply to be $36$ by constructing a table of factorizations of $36.$
$$
\begin{array}{ll|l}
\displaystyle &\displaystyle\mbox{Factors} &\mbox{Sum}\\\hline
\displaystyle 36 &\displaystyle= 1\cdot 36 & 37\mbox{}\\
\displaystyle &\displaystyle= 2\cdot 18 & 20\mbox{}\\
\displaystyle &\displaystyle= 3\cdot 12 & 15\mbox{}\\
\displaystyle &\displaystyle= 4\cdot 9 & 13\, \mbox{ almost!}\\
\displaystyle &\displaystyle= (-4)\cdot (-9) & -13 \, \checkmark \mbox{}\\
\displaystyle &\displaystyle= 6\cdot 6 & 12 \mbox{}\\
\end{array}
$$
Since $-4$ and $-9$ add to $-13$ and multiply to be $36,$ we have our factorization.
$$q^2-13q+36=(q-4)(q-9)$$
Check
$$ \begin{array}{lll} \displaystyle (q-4)(q-9)&\displaystyle=q^2-9q-4q+36 &\mbox{}\\ \displaystyle &\displaystyle=q^2-13q+36 & \checkmark \mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle (q-4)(q-9)&\displaystyle=q^2-9q-4q+36 &\mbox{}\\ \displaystyle &\displaystyle=q^2-13q+36 & \checkmark \mbox{}\\ \end{array} $$
This time, we'll skip the table.
Since $-4$ and $12$ add to $8$ and multiply to be $-48,$ we have our factorization. $$p^2+8p-48=(p-4)(p+12)$$ Check
$$ \begin{array}{lll} \displaystyle(p-4)(p+12) &\displaystyle=p^2+12p-4p-48 &\mbox{}\\ \displaystyle &\displaystyle=p^2+8p-48 & \checkmark \mbox{}\\ \end{array} $$
Since $-4$ and $12$ add to $8$ and multiply to be $-48,$ we have our factorization. $$p^2+8p-48=(p-4)(p+12)$$ Check
$$ \begin{array}{lll} \displaystyle(p-4)(p+12) &\displaystyle=p^2+12p-4p-48 &\mbox{}\\ \displaystyle &\displaystyle=p^2+8p-48 & \checkmark \mbox{}\\ \end{array} $$
$x^2+2x+12$
We're looking for two factors which add to $2$ and multiply to $12.$
Looking at all the possible factors of $12,$ $$ \begin{array}{ll|l} \displaystyle &\displaystyle\mbox{Factors} &\mbox{Sum}\\\hline \displaystyle 12 &\displaystyle= 1\cdot 12 & 13\mbox{}\\ \displaystyle &\displaystyle= 2\cdot 6 & 8\mbox{}\\ \displaystyle &\displaystyle= 3\cdot 4 & 7\mbox{}\\ \end{array} $$ Since none of the possible factorizations of $12$ add up to $2,$ we conclude that $x^2+2x+12$ is not factorable.
In this case we may say that $x^2+2x+12$ is a prime polynomial.
Looking at all the possible factors of $12,$ $$ \begin{array}{ll|l} \displaystyle &\displaystyle\mbox{Factors} &\mbox{Sum}\\\hline \displaystyle 12 &\displaystyle= 1\cdot 12 & 13\mbox{}\\ \displaystyle &\displaystyle= 2\cdot 6 & 8\mbox{}\\ \displaystyle &\displaystyle= 3\cdot 4 & 7\mbox{}\\ \end{array} $$ Since none of the possible factorizations of $12$ add up to $2,$ we conclude that $x^2+2x+12$ is not factorable.
In this case we may say that $x^2+2x+12$ is a prime polynomial.
Another Common Form: $x^2+bxy+cy^2$
Fact: Expressions of the form $x^2+bxy+cy^2$ factor into $$(x+ry)(x+sy).$$
Multiplying this expression out we get $x^2+(r+s)xy+rsy^2.$
So again, factoring these kinds of expressions amounts to finding two numbers whose sum is $b$ and whose product is $c$.
Examples
$b^2+9b v+14v^2$
We can factor this as if the $v$ wasn't even there.
That is, we can find two numbers which add to $9$ and multiply to $14,$ we can factor the expression as $$ (b+\mbox{___}v)(b+\mbox{___}v) $$ where the two numbers we find go into the blanks.
Now, skipping the table, two numbers which add to be $9$ and multiply to be $14$ are $2$ and $7.$
Our factorization is then $$ (b+2v)(b+7v) $$ Check
$$ \begin{array}{lll} \displaystyle (b+2v)(b+7v)&\displaystyle=b^2+7bv+2vb+14v^2 &\mbox{}\\ \displaystyle &\displaystyle=b^2+7bv+2bv+14v^2 & \mbox{reorder $vb$ as $bv$}\\ \displaystyle &\displaystyle=b^2+9bv+14v^2 \,\checkmark & \mbox{combine like terms}\\ \end{array} $$
That is, we can find two numbers which add to $9$ and multiply to $14,$ we can factor the expression as $$ (b+\mbox{___}v)(b+\mbox{___}v) $$ where the two numbers we find go into the blanks.
Now, skipping the table, two numbers which add to be $9$ and multiply to be $14$ are $2$ and $7.$
Our factorization is then $$ (b+2v)(b+7v) $$ Check
$$ \begin{array}{lll} \displaystyle (b+2v)(b+7v)&\displaystyle=b^2+7bv+2vb+14v^2 &\mbox{}\\ \displaystyle &\displaystyle=b^2+7bv+2bv+14v^2 & \mbox{reorder $vb$ as $bv$}\\ \displaystyle &\displaystyle=b^2+9bv+14v^2 \,\checkmark & \mbox{combine like terms}\\ \end{array} $$
$\beta^2-11\beta r+30r^2$
$-5$ and $-6$ add to $-11$ and multiply to $30.$ So,
$$
\beta^2-11\beta r+30r^2=(\beta -5r)(\beta -6r)
$$
Check
$$ \begin{array}{lll} \displaystyle(\beta -5r)(\beta -6r) &\displaystyle=\beta^2-6\beta r -5\beta r +30 r^2&\mbox{}\\ \displaystyle &\displaystyle= \beta^2-11\beta r+30r^2& \checkmark \mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle(\beta -5r)(\beta -6r) &\displaystyle=\beta^2-6\beta r -5\beta r +30 r^2&\mbox{}\\ \displaystyle &\displaystyle= \beta^2-11\beta r+30r^2& \checkmark \mbox{}\\ \end{array} $$
$s^2-5st-50t^2$
$5$ and $-10$ add to $-5$ and multiply to $-50.$ So,
$$
s^2-5st-50t^2=(s+5t)(s-10t)
$$
Check
$$ \begin{array}{lll} \displaystyle (s+5t)(s-10t)&\displaystyle=s^2-10st+5st-50t^2 &\mbox{}\\ \displaystyle &\displaystyle= s^2-5st-50t^2& \checkmark \mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle (s+5t)(s-10t)&\displaystyle=s^2-10st+5st-50t^2 &\mbox{}\\ \displaystyle &\displaystyle= s^2-5st-50t^2& \checkmark \mbox{}\\ \end{array} $$
Other Variarions : Completely factor the following polynomials.
$-h^2+10h-21$
We first factor out a negative so that the squared term has a $1$ in front of it.
$$
-h^2+10h-21=-(h^2-10h+21)
$$
We may now proceed as usual: find two numbers which add to $-10$ and multiply to $21.$
Since $-3$ and $-7$ add to $-10$ and multiply to $21,$ the full calculation is, $$ \begin{array}{lll} \displaystyle -h^2+10h-21&\displaystyle=-(h^2-10h+21) &\mbox{factor out minus}\\ \displaystyle &\displaystyle=-(h-3)(h-7) &\mbox{}\\ \end{array} $$ Check
$$ \begin{array}{lll} \displaystyle -(h-3)(h-7) &\displaystyle=-(h^2-7h-3h+21) &\mbox{multiply binomial factors}\\ \displaystyle &\displaystyle=-(h^2-10h+21) &\mbox{combine like terms}\\ \displaystyle &\displaystyle= -h^2+10h-21 \checkmark & \mbox{distribute minus}\\ \end{array} $$
Since $-3$ and $-7$ add to $-10$ and multiply to $21,$ the full calculation is, $$ \begin{array}{lll} \displaystyle -h^2+10h-21&\displaystyle=-(h^2-10h+21) &\mbox{factor out minus}\\ \displaystyle &\displaystyle=-(h-3)(h-7) &\mbox{}\\ \end{array} $$ Check
$$ \begin{array}{lll} \displaystyle -(h-3)(h-7) &\displaystyle=-(h^2-7h-3h+21) &\mbox{multiply binomial factors}\\ \displaystyle &\displaystyle=-(h^2-10h+21) &\mbox{combine like terms}\\ \displaystyle &\displaystyle= -h^2+10h-21 \checkmark & \mbox{distribute minus}\\ \end{array} $$
$tc^3-17tc^2+70tc$
Let's not forget that we can factor out GCFs!
$$
\begin{array}{lll}
\displaystyle tc^3-17tc^2+70tc&\displaystyle=tc(c^2-17c+70) &\mbox{factor out GCF}\\
\displaystyle &\displaystyle= tc(c-7)(c-10)&\mbox{factor trinomial}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle tc(c-7)(c-10)&\displaystyle=tc(c^2-10c-7c+70) &\mbox{multiply binomial factors}\\ \displaystyle &\displaystyle=tc(c^2-17c+70) & \mbox{combine like terms}\\ \displaystyle &\displaystyle= tc^3-17tc^2+70tc \, \checkmark & \mbox{distribute $tc$}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle tc(c-7)(c-10)&\displaystyle=tc(c^2-10c-7c+70) &\mbox{multiply binomial factors}\\ \displaystyle &\displaystyle=tc(c^2-17c+70) & \mbox{combine like terms}\\ \displaystyle &\displaystyle= tc^3-17tc^2+70tc \, \checkmark & \mbox{distribute $tc$}\\ \end{array} $$
$sr^3+7r^2 s^2+12rs^3$
$$
\begin{array}{lll}
\displaystyle sr^3+7r^2 s^2+12rs^3 &\displaystyle=rs(r^2+7rs+12s^2) &\mbox{factor out GCF}\\
\displaystyle &\displaystyle=rs(r+3s)(r+4s) &\mbox{factor trinomial}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle rs(r+3s)(r+4s)&\displaystyle=rs(r^2+4rs+3rs+12s^2) &\mbox{multiply binomial factors}\\ \displaystyle &\displaystyle=rs(r^2+7rs+12s^2) &\mbox{combine like terms}\\ \displaystyle &\displaystyle=sr^3+7r^2s^2+12rs^3 \, \checkmark & \mbox{distribute $rs$}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle rs(r+3s)(r+4s)&\displaystyle=rs(r^2+4rs+3rs+12s^2) &\mbox{multiply binomial factors}\\ \displaystyle &\displaystyle=rs(r^2+7rs+12s^2) &\mbox{combine like terms}\\ \displaystyle &\displaystyle=sr^3+7r^2s^2+12rs^3 \, \checkmark & \mbox{distribute $rs$}\\ \end{array} $$