ORCCA Factoring Trinomials with Leading Coefficient One

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Section 10.3 Factoring Trinomials with Leading Coefficient One

Objectives: PCC Course Content and Outcome Guide

MTH 95 CCOG 2.c

In Chapter 5, we learned how to multiply binomials like $(x+2)(x+3)$ and obtain the trinomial $x^2+5x+6\text{.}$ In this section, we will learn how to undo that. So we'll be starting with a trinomial like $x^2+5x+6$ and obtaining its factored form $(x+2)(x+3)\text{.}$ The trinomials that we'll factor in this section all have leading coefficient $1\text{,}$ but Section 10.4 will cover some more general trinomials.

Figure 10.3.1. Alternative Video Lesson

Subsection 10.3.1 Factoring Trinomials by Listing Factor Pairs

Consider the example $x^2+\firsthighlight{5}x+\secondhighlight{6}=(x+2)(x+3)\text{.}$ There are at least three things that are important to notice:

The leading coefficient of $x^2+\firsthighlight{5}x+\secondhighlight{6}$ is $1\text{.}$
The two factors on the right use the numbers $2$ and $3\text{,}$ and when you multiply these you get the $\secondhighlight{6}\text{.}$
The two factors on the right use the numbers $2$ and $3\text{,}$ and when you add these you get the $\firsthighlight{5}\text{.}$

So the idea is that if you need to factor $x^2+5x+6$ and you somehow discover that $2$ and $3$ are special numbers (because $2\cdot3=6$ and $2+3=5$ ), then you can conclude that $(x+2)(x+3)$ is the factored form of the given polynomial.

Example 10.3.2.

Factor $x^2+13x+40\text{.}$ Since the leading coefficient is $1\text{,}$ we are looking to write this polynomial as $(x+\mathord{?})(x+\mathord{?})$ where the question marks are two possibly different, possibly negative, numbers. We need these two numbers to multiply to $40$ and add to $13\text{.}$ How can you track these two numbers down? Since the numbers need to multiply to $40\text{,}$ one method is to list all factor pairs of $40$ in a table just to see what your options are. We'll write every pair of factors that multiply to $40\text{.}$

1 \cdot 40

$1\cdot40$

2 \cdot 20

$2\cdot20$

4 \cdot 10

$4\cdot10$

5 \cdot 8

$5\cdot8$

- 1 \cdot (- 40)

$-1\cdot(-40)$

- 2 \cdot (- 20)

$-2\cdot(-20)$

- 4 \cdot (- 10)

$-4\cdot(-10)$

- 5 \cdot (- 8)

$-5\cdot(-8)$

We wanted to find all factor pairs. To avoid missing any, we started using $1$ as a factor, and then slowly increased that first factor. The table skips over using $3$ as a factor, because $3$ is not a factor of $40\text{.}$ Similarly the table skips using $6$ and $7$ as a factor. And there would be no need to continue with $8$ and beyond, because we already found “large” factors like $8$ as the partners of “small” factors like $5\text{.}$

There is an entire second column where the signs are reversed, since these are also ways to multiply two numbers to get $40\text{.}$ In the end, there are eight factor pairs.

We need a pair of numbers that also adds to $13\text{.}$ So we check what each of our factor pairs add up to:

Factor Pair	Sum of the Pair
$1\cdot40$	$41$
$2\cdot20$	$22$
$4\cdot10$	$14$
$5\cdot8$	$13$ (what we wanted)

Factor Pair	Sum of the Pair
$-1\cdot(-40)$	(no need to go this far)
$-2\cdot(-20)$	(no need to go this far)
$-4\cdot(-10)$	(no need to go this far)
$-5\cdot(-8)$	(no need to go this far)

The winning pair of numbers is $5$ and $8\text{.}$ Again, what matters is that $5\cdot8=40\text{,}$ and $5+8=13\text{.}$ So we can conclude that $x^2+13x+40=(x+5)(x+8)\text{.}$

permalinkTo ensure that we made no mistakes, here are some possible checks.

Multiply it Out.

Multiplying out our answer $(x+5)(x+8)$ should give us $x^2+13x+40\text{.}$

\begin{aligned} (x + 5) (x + 8) & = (x + 5) \cdot x + (x + 5) \cdot 8 \\ = x^{2} + 5 x + 8 x + 40 \\ \overset{✓}{=} x^{2} + 13 x + 40 \end{aligned}

$\begin{align*} (x+5)(x+8)\amp=(x+5)\cdot x+(x+5)\cdot8\\ \amp=x^2+5x+8x+40\\ \amp\stackrel{\checkmark}{=}x^2+13x+40 \end{align*}$

We could also use a rectangular area diagram to verify the factorization is correct:

	$x$	$5$
$x$	$x^2$	$5x$
$8$	$8x$	$40$

Evaluating.

If the answer really is $(x+5)(x+8)\text{,}$ then notice how evaluating at $-5$ would result in $0\text{.}$ So the original expression should also result in $0$ if we evaluate at $\substitute{-5}\text{.}$ And similarly, if we evaluate it at $\substitute{-8}\text{,}$ $x^2+13x+40$ should be $0\text{.}$

\begin{aligned} (- 5)^{2} + 13 (- 5) + 40 & \overset{?}{=} 0 & (- 8)^{2} + 13 (- 8) + 40 & \overset{?}{=} 0 \\ 25 - 65 + 40 & \overset{?}{=} 0 & 64 - 104 + 40 & \overset{?}{=} 0 \\ 0 & \overset{✓}{=} 0 & 0 & \overset{✓}{=} 0 . \end{aligned}

$\begin{align*} (\substitute{-5})^2+13(\substitute{-5})+40\amp\stackrel{?}{=}0\amp(\substitute{-8})^2+13(\substitute{-8})+40\amp\stackrel{?}{=}0\\ 25-65+40\amp\stackrel{?}{=}0\amp 64-104+40\amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0\amp 0\amp\stackrel{\checkmark}{=}0\text{.} \end{align*}$

This also gives us evidence that the factoring was correct.

Example 10.3.3.

Factor $y^2-11y+24\text{.}$ The negative coefficient is a small complication from Example 10.3.2, but the process is actually still the same.

We need a pair of numbers that multiply to \(24\) and add to \(-11\text{.}\) Note that we do care to keep track that they sum to a negative total.

Factor Pair	Sum of the Pair
\(1\cdot24\)	\(25\)
\(2\cdot12\)	\(14\)
\(3\cdot8\)	\(11\) (close; wrong sign)
\(4\cdot6\)	\(10\)

Factor Pair	Sum of the Pair
\(-1\cdot(-24)\)	\(-25\)
\(-2\cdot(-12)\)	\(-14\)
\(-3\cdot(-8)\)	\(-11\) (what we wanted)
\(-4\cdot(-6)\)	(no need to go this far)

So \(y^2-11y+24=(y-3)(y-8)\text{.}\) To confirm that this is correct, we should check. Either by multiplying out the factored form:

\begin{align*} (y-3)(y-8)\amp=(y-3)\cdot y-(y-3)\cdot8\\ \amp=y^2-3y-8y+24\\ \amp\stackrel{\checkmark}{=}y^2-11y+24 \end{align*}

	\(y\)	\(-3\)
\(y\)	\(y^2\)	\(-3y\)
\(-8\)	\(-8y\)	\(24\)

Or by evaluating the original expression at \(\substitute{3}\) and \(\substitute{8}\text{:}\)

\begin{align*} \substitute{3}^2-11(\substitute{3})+24\amp\stackrel{?}{=}0\amp\substitute{8}^2-11(\substitute{8})+24\amp\stackrel{?}{=}0\\ 9-33+24\amp\stackrel{?}{=}0\amp 64-88+24\amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0\amp 0\amp\stackrel{\checkmark}{=}0\text{.} \end{align*}

Our factorization passes the tests.

Example 10.3.4.

Factor $z^2+5z-6\text{.}$ The negative coefficient is again a small complication from Example 10.3.2, but the process is actually still the same.

We need a pair of numbers that multiply to \(-6\) and add to \(5\text{.}\) Note that we do care to keep track that they multiply to a negative product.

Factor Pair	Sum of the Pair
\(1\cdot(-6)\)	\(-5\) (close; wrong sign)
\(2\cdot(-3)\)	\(14\)

Factor Pair	Sum of the Pair
\(-1\cdot6\)	\(5\) (what we wanted)
\(-2\cdot3\)	(no need to go this far)

So \(z^2+5z-6=(z-1)(z+6)\text{.}\) To confirm that this is correct, we should check. Either by multiplying out the factored form:

\begin{align*} (z-1)(z+6)\amp=(z-1)\cdot z+(z-1)\cdot6\\ \amp=z^2-z+6z-6\\ \amp\stackrel{\checkmark}{=}z^2+5z-6 \end{align*}

	\(z\)	\(-1\)
\(z\)	\(z^2\)	\(-z\)
\(6\)	\(6z\)	\(-6\)

Or by evaluating the original expression at \(\substitute{1}\) and \(\substitute{-6}\text{:}\)

\begin{align*} \substitute{1}^2+5(\substitute{1})-6\amp\stackrel{?}{=}0\amp(\substitute{-6})^2+5(\substitute{-6})-6\amp\stackrel{?}{=}0\\ 1+5-6\amp\stackrel{?}{=}0\amp 36-30-6\amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0\amp 0\amp\stackrel{\checkmark}{=}0\text{.} \end{align*}

Our factorization passes the tests.

Checkpoint 10.3.5.

Subsection 10.3.2 Connection to Grouping

permalinkThe factoring method we just learned is actually taking a shortcut compared to what we will learn in Section 10.4. To prepare yourself for that more complicated factoring technique, you may want to try taking the “scenic route” instead of that shortcut.

Example 10.3.6.

Let's factor $x^2+13x+40$ again (the polynomial from Example 10.3.2). As before, it is important to discover that $5$ and $8$ are important numbers, because they multiply to $40$ and add to $13\text{.}$ As before, listing out all of the factor pairs is one way to discover the $5$ and the $8\text{.}$

Instead of jumping to the factored answer, we can show how $x^2+13x+40$ factors in a more step-by-step fashion using $5$ and $8\text{.}$ Since they add up to $13\text{,}$ we can write:

\begin{aligned} x^{2} + \overset{↓}{13} x + 40 & = x^{2} + \overset{↓}{\overset{⏞}{5 x + 8 x}} + 40 \end{aligned}

$\begin{align*} x^2+\attention{13}x+40\amp=x^2+\attention{\overbrace{5x+8x}}+40\\ \end{align*}$

We have intentionally split up the trinomial into an unsimplified polynomial with four terms. In Section 10.2, we handled such four-term polynomials by grouping:

\begin{aligned} = (x^{2} + 5 x) + (8 x + 40) \end{aligned}

$\begin{align*} \amp=\left(x^2+5x\right)+(8x+40)\\ \end{align*}$

Now we can factor out each group's greatest common factor:

\begin{aligned} = x (x + 5) + 8 (x + 5) \\ = x \overset{↓}{\overset{⏞}{(x + 5)}} + 8 \overset{↓}{\overset{⏞}{(x + 5)}} \\ = (x + 5) (x + 8) \end{aligned}

$\begin{align*} \amp=x(x+5)+8(x+5)\\ \amp=x\attention{\overbrace{(x+5)}}+8\attention{\overbrace{(x+5)}}\\ \amp=(x+5)(x+8) \end{align*}$

And we have found that $x^2+13x+40$ factors as $(x+5)(x+8)$ without taking the shortcut.

permalinkThis approach takes more time, and ultimately you may not use it much. However, if you try a few examples this way, it may make you more comfortable with the more complicated technique in Section 10.4.

Subsection 10.3.3 Trinomials with Higher Powers

So far we have only factored examples of quadratic trinomials: trinomials whose highest power of the variable is $2\text{.}$ However, this technique can also be used to factor trinomials where there is a larger highest power of the variable. It only requires that the highest power is even, that the next highest power is half of the highest power, and that the third term is a constant term.

permalinkIn the four examples below, check:

if the highest power is even
if the next highest power is half of the highest power
if the last term is constant

Factor pairs will help with…

$y^6-23y^3-50$
$h^{16}+22h^8+105$

Factor pairs won't help with…

$y^5-23y^3-50$
$h^{16}+22h^8+105h^2$

Example 10.3.7.

Factor $h^{16}+22h^8+105\text{.}$ This polynomial is one of the examples above where using factor pairs will help. We find that $7\cdot15=105\text{,}$ and $7+15=22\text{,}$ so the numbers $7$ and $15$ can be used:

\begin{aligned} h^{16} + 22 h^{8} + 105 & = h^{16} + \overset{⏞}{7 h^{8} + 15 h^{8}} + 105 \\ = (h^{16} + 7 h^{8}) + (15 h^{8} + 105) \\ = h^{8} (h^{8} + 7) + 15 (h^{8} + 7) \\ = (h^{8} + 7) (h^{8} + 15) \end{aligned}

$\begin{align*} h^{16}+22h^8+105\amp=h^{16}+\overbrace{7h^8+15h^8}+105\\ \amp=\left(h^{16}+7h^8\right)+\left(15h^8+105\right)\\ \amp=h^8\left(h^8+7\right)+15\left(h^8+7\right)\\ \amp=\left(h^8+7\right)\left(h^8+15\right) \end{align*}$

Actually, once we settled on using $7$ and $15\text{,}$ we could have concluded that $h^{16}+22h^8+105$ factors as $\left(h^8+7\right)\left(h^8+15\right)\text{,}$ if we know which power of $h$ to use. We'll always use half the highest power in these factorizations.

In any case, to confirm that this is correct, we should check by multiplying out the factored form:

$\begin{aligned}[t] (h^8+7)(h^8+15)\amp=(h^8+7)\cdot h^8+(h^8+7)\cdot15\\ \amp=h^{16}+7h^8+15h^8+105\\ \amp\stackrel{\checkmark}{=}h^{16}+22h^8+15\end{aligned}$

	$h^8$	$7$
$h^8$	$h^{16}$	$7h^8$
$15$	$15h^8$	$105$

Our factorization passes the tests.

Checkpoint 10.3.8.

Subsection 10.3.4 Factoring in Stages

permalinkSometimes factoring a polynomial will take two or more “stages.” Always begin factoring a polynomial by factoring out its greatest common factor, and then apply a second stage where you use a technique from this section. The process of factoring a polynomial is not complete until each of the factors cannot be factored further.

Example 10.3.9.

Factor $2z^2-6z-80\text{.}$

We will first factor out the common factor, \(2\text{:}\)

\begin{equation*} 2z^2-6z-80=2\left(z^2-3z-40\right) \end{equation*}

Now we are left with a factored expression that might factor more. Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be \(-40\) and add to be \(-3\text{?}\)” Since \(5\) and \(-8\) do the job the full factorization is:

\begin{align*} 2z^2-6z-80\amp=2\left(z^2-3z-40\right)\\ \amp=2(z+5)(z-8) \end{align*}

Example 10.3.10.

Factor $-r^2+2r+24\text{.}$

The three terms don't exactly have a common factor, but as discussed in Section 10.1, when the leading term has a negative sign, it is often helpful to factor out that negative sign:

\begin{equation*} -r^2+2r+24=-\mathopen{}\left(r^2-2r-24\right)\mathclose{}\text{.} \end{equation*}

Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be \(-24\) and add to be \(-2\text{?}\)” Since \(-6\) and \(4\) work here and the full factorization is shown:

\begin{align*} -r^2+2r+24\amp=-\mathopen{}\left(r^2-2r-24\right)\mathclose{}\\ \amp=-(r-6)(r+4) \end{align*}

Example 10.3.11.

Factor $p^2q^3+4p^2q^2-60p^2q\text{.}$

First, always look for the greatest common factor: in this trinomial it is \(p^2q\text{.}\) After factoring this out, we have

\begin{equation*} p^2q^3+4p^2q^2-60p^2q=p^2q\mathopen{}\left(q^2+4q-60\right)\mathclose{}\text{.} \end{equation*}

Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be \(-60\) and add to be \(4\text{?}\)” Since \(10\) and \(-6\) fit the bill, the full factorization can be shown below:

\begin{align*} p^2q^3+4p^2q^2-60p^2q\amp=p^2q\mathopen{}\left(q^2+4q-60\right)\mathclose{}\\ \amp=p^2q(q+10)(q-6) \end{align*}

Subsection 10.3.5 More Trinomials with Two Variables

You might encounter a trinomial with two variables that can be factored using the methods we've discussed in this section. It can be tricky though: $x^2+5xy+6y^2$ has two variables and it can factor using the methods from this section, but $x^2+5x+6y^2$ also has two variables and it cannot be factored. So in examples of this nature, it is even more important to check that factorizations you find actually work.

Example 10.3.12.

Factor $x^2+5xy+6y^2\text{.}$ This is a trinomial, and the coefficient of $x$ is $1\text{,}$ so maybe we can factor it. We want to write $(x+\mathord{?})(x+\mathord{?})$ where the question marks will be something that makes it all multiply out to $x^2+5xy+6y^2\text{.}$

Since the last term in the polynomial has a factor of $y^2\text{,}$ it is natural to wonder if there is a factor of $y$ in each of the two question marks. If there were, these two factors of $y$ would multiply to $y^2\text{.}$ So it is natural to wonder if we are looking for $(x+\mathord{?}y)(x+\mathord{?}y)$ where now the question marks are just numbers.

At this point we can think like we have throughout this section. Are there some numbers that multiply to $6$ and add to $5\text{?}$ Yes, specifically $2$ and $3\text{.}$ So we suspect that $(x+2y)(x+3y)$ might be the factorization.

To confirm that this is correct, we should check by multiplying out the factored form:

\begin{aligned} (x + 2 y) (x + 3 y) & = (x + 2 y) \cdot x + (x + 2 y) \cdot 3 y \\ = x^{2} + 2 x y + 3 x y + 6 y^{2} \\ \overset{✓}{=} x^{2} + 5 x y + 6 y^{2} \end{aligned}

$\begin{align*} (x+2y)(x+3y)\amp=(x+2y)\cdot x+(x+2y)\cdot3y\\ \amp=x^2+2xy+3xy+6y^2\\ \amp\stackrel{\checkmark}{=}x^2+5xy+6y^2 \end{align*}$

	$x$	$2y$
$x$	$x^2$	$2xy$
$3y$	$3xy$	$6y^2$

Our factorization passes the tests.

permalinkIn Section 10.4, there is a more definitive method for factoring polynomials of this form.

Reading Questions 10.3.6 Reading Questions

1.

To factor $x^2+bx+c\text{,}$ you look for two numbers that do what?

2.

How many factor pairs are there for the number $6\text{?}$

Exercises 10.3.7 Exercises

Review and Warmup

1.

Multiply the polynomials.

$\left({y+5}\right)\left({y+10}\right)=$

2.

Multiply the polynomials.

$\left({r+1}\right)\left({r+4}\right)=$

3.

Multiply the polynomials.

$\left({r+8}\right)\left({r-3}\right)=$

4.

Multiply the polynomials.

$\left({r+4}\right)\left({r-8}\right)=$

5.

Multiply the polynomials.

$\left({t-10}\right)\left({t-4}\right)=$

6.

Multiply the polynomials.

$\left({t-4}\right)\left({t-10}\right)=$

7.

Multiply the polynomials.

${3\!\left(x+2\right)\!\left(x+3\right)}=$

8.

Multiply the polynomials.

${-4\!\left(y-1\right)\!\left(y-9\right)}=$

9.

Multiply the polynomials.

${2\!\left(y-10\right)\!\left(y-3\right)}=$

10.

Multiply the polynomials.

${-2\!\left(r+7\right)\!\left(r+6\right)}=$

Factoring Trinomials with Leading Coefficient One

11.

Factor the given polynomial.

${r^{2}+12r+20}=$

12.

Factor the given polynomial.

${r^{2}+13r+42}=$

13.

Factor the given polynomial.

${r^{2}+13r+30}=$

14.

Factor the given polynomial.

${t^{2}+14t+40}=$

15.

Factor the given polynomial.

${t^{2}+3t-18}=$

16.

Factor the given polynomial.

${x^{2}+7x-30}=$

17.

Factor the given polynomial.

${x^{2}+2x-63}=$

18.

Factor the given polynomial.

${y^{2}-y-30}=$

19.

Factor the given polynomial.

${y^{2}-7y+10}=$

20.

Factor the given polynomial.

${r^{2}-14r+45}=$

21.

Factor the given polynomial.

${r^{2}-8r+15}=$

22.

Factor the given polynomial.

${r^{2}-10r+16}=$

23.

Factor the given polynomial.

${t^{2}+10t+16}=$

24.

Factor the given polynomial.

${t^{2}+11t+30}=$

25.

Factor the given polynomial.

${x^{2}+11x+10}=$

26.

Factor the given polynomial.

${x^{2}+12x+32}=$

27.

Factor the given polynomial.

${y^{2}-6y-40}=$

28.

Factor the given polynomial.

${y^{2}+6y-7}=$

29.

Factor the given polynomial.

${r^{2}+3r-28}=$

30.

Factor the given polynomial.

${r^{2}-3r-4}=$

31.

Factor the given polynomial.

${r^{2}-6r+5}=$

32.

Factor the given polynomial.

${t^{2}-16t+63}=$

33.

Factor the given polynomial.

${t^{2}-7t+12}=$

34.

Factor the given polynomial.

${x^{2}-18x+80}=$

35.

Factor the given polynomial.

${x^{2}+x+6}=$

36.

Factor the given polynomial.

${y^{2}+3y+10}=$

37.

Factor the given polynomial.

${y^{2}+9}=$

38.

Factor the given polynomial.

${r^{2}-4r+6}=$

39.

Factor the given polynomial.

${r^{2}+6r+9}=$

40.

Factor the given polynomial.

${r^{2}+22r+121}=$

41.

Factor the given polynomial.

${t^{2}+14t+49}=$

42.

Factor the given polynomial.

${t^{2}+6t+9}=$

43.

Factor the given polynomial.

${x^{2}-20x+100}=$

44.

Factor the given polynomial.

${x^{2}-12x+36}=$

45.

Factor the given polynomial.

${y^{2}-4y+4}=$

46.

Factor the given polynomial.

${y^{2}-20y+100}=$

47.

Factor the given polynomial.

${2r^{2}-2r-40}=$

48.

Factor the given polynomial.

${2r^{2}-2r-4}=$

49.

Factor the given polynomial.

${10r^{2}-10}=$

50.

Factor the given polynomial.

${2t^{2}+6t-20}=$

51.

Factor the given polynomial.

${10t^{2}-30t+20}=$

52.

Factor the given polynomial.

${2x^{2}-18x+16}=$

53.

Factor the given polynomial.

${3x^{2}-15x+12}=$

54.

Factor the given polynomial.

${3y^{2}-21y+18}=$

55.

Factor the given polynomial.

${2y^{6}+20y^{5}+18y^{4}}=$

56.

Factor the given polynomial.

${2y^{7}+14y^{6}+12y^{5}}=$

57.

Factor the given polynomial.

${3r^{6}+24r^{5}+21r^{4}}=$

58.

Factor the given polynomial.

${5r^{7}+20r^{6}+15r^{5}}=$

59.

Factor the given polynomial.

${10t^{7}-20t^{6}-30t^{5}}=$

60.

Factor the given polynomial.

${9t^{4}+9t^{3}-18t^{2}}=$

61.

Factor the given polynomial.

${3x^{10}-12x^{8}}=$

62.

Factor the given polynomial.

${5x^{6}-20x^{5}-25x^{4}}=$

63.

Factor the given polynomial.

${2y^{7}-8y^{6}+6y^{5}}=$

64.

Factor the given polynomial.

${5y^{5}-15y^{4}+10y^{3}}=$

65.

Factor the given polynomial.

${3y^{7}-12y^{6}+9y^{5}}=$

66.

Factor the given polynomial.

${6r^{4}-18r^{3}+12r^{2}}=$

67.

Factor the given polynomial.

${-r^{2}-4r+45}=$

68.

Factor the given polynomial.

${-t^{2}+4t+12}=$

69.

Factor the given polynomial.

${-t^{2}+5t+24}=$

70.

Factor the given polynomial.

${-x^{2}-4x+5}=$

71.

Factor the given polynomial.

${x^{2}+12xr+20r^{2}}=$

72.

Factor the given polynomial.

${y^{2}+7yr+6r^{2}}=$

73.

Factor the given polynomial.

${y^{2}-3yt-10t^{2}}=$

74.

Factor the given polynomial.

${y^{2}-4yx-32x^{2}}=$

75.

Factor the given polynomial.

${r^{2}-7rt+12t^{2}}=$

76.

Factor the given polynomial.

${r^{2}-5ry+4y^{2}}=$

77.

Factor the given polynomial.

${t^{2}+16tx+64x^{2}}=$

78.

Factor the given polynomial.

${t^{2}+2tr+r^{2}}=$

79.

Factor the given polynomial.

${x^{2}-12xy+36y^{2}}=$

80.

Factor the given polynomial.

${x^{2}-22xt+121t^{2}}=$

81.

Factor the given polynomial.

${4y^{2}+20y+16}=$

82.

Factor the given polynomial.

${2y^{2}+18y+16}=$

83.

Factor the given polynomial.

${3x^{2}y+15xy+18y}=$

84.

Factor the given polynomial.

${2x^{2}y+6xy+4y}=$

85.

Factor the given polynomial.

${7a^{2}b-7ab-14b}=$

86.

Factor the given polynomial.

${10a^{2}b+10ab-20b}=$

87.

Factor the given polynomial.

${2x^{2}y-22xy+20y}=$

88.

Factor the given polynomial.

${9x^{2}y-27xy+18y}=$

89.

Factor the given polynomial.

${3x^{3}y+24x^{2}y+21xy}=$

90.

Factor the given polynomial.

${2x^{3}y+16x^{2}y+30xy}=$

91.

Factor the given polynomial.

${x^{2}y^{2}+x^{2}yz-20x^{2}z^{2}}=$

92.

Factor the given polynomial.

${x^{2}y^{2}+8x^{2}yz-9x^{2}z^{2}}=$

93.

Factor the given polynomial.

${r^{2}+0.9r+0.2}=$

94.

Factor the given polynomial.

${r^{2}+0.9r+0.14}=$

95.

Factor the given polynomial.

${t^{2}x^{2}+6tx+5}=$

96.

Factor the given polynomial.

${t^{2}y^{2}+5ty+6}=$

97.

Factor the given polynomial.

${x^{2}t^{2}-2xt-24}=$

98.

Factor the given polynomial.

${x^{2}r^{2}+xr-20}=$

99.

Factor the given polynomial.

${y^{2}t^{2}-8yt+7}=$

100.

Factor the given polynomial.

${y^{2}x^{2}-9yx+14}=$

101.

Factor the given polynomial.

${6y^{2}r^{2}+18yr+12}=$

102.

Factor the given polynomial.

${5r^{2}y^{2}+15ry+10}=$

103.

Factor the given polynomial.

${7r^{2}t^{2}-7}=$

104.

Factor the given polynomial.

${3t^{2}r^{2}+6tr-24}=$

105.

Factor the given polynomial.

${6x^{2}y^{3}-18xy^{2}+12y}=$

106.

Factor the given polynomial.

${6x^{2}y^{3}-24xy^{2}+18y}=$

107.

Factor the given polynomial.

${\left(a+b\right)x^{2}+8\!\left(a+b\right)x+12\!\left(a+b\right)}=$

108.

Factor the given polynomial.

${\left(a+b\right)y^{2}+9\!\left(a+b\right)y+18\!\left(a+b\right)}=$

Challenge

109.

What integers can go in the place of $b$ so that the quadratic expression ${x^{2}+bx+10}$ is factorable?