Factoring Monic Quadratics with Integer Factors

Welcome back. In the last lesson, you expanded products such as

$$(x+2)(x+4)$$

to get

$$x^2+6x+8.$$

This lesson reverses that process. Given a monic quadratic trinomial—a quadratic whose $x^2$ coefficient is $1$—you will rewrite it as a product of two binomials. This is called factoring. It is a central algebra skill for graphing because the factors will soon reveal where a parabola crosses the $x$-axis.

By the end, you should be able to factor expressions of the form

$$x^2+bx+c$$

whenever integer factors exist.

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Factoring is undoing expansion

Last lesson established the pattern

$$(x+p)(x+q)=x^2+(p+q)x+pq.$$

The two numbers inside the parentheses, $p$ and $q$, play two roles:

• Their sum becomes the coefficient of $x$.
• Their product becomes the constant term.

Factoring asks you to work backward. For

$$x^2+bx+c,$$

find integers $p$ and $q$ such that

$$p+q=b$$

and

$$pq=c.$$

Then the factored expression is

$$\boxed{(x+p)(x+q)}.$$

This shortcut works specifically because the leading term is $x^2$, with coefficient $1$. That is what monic means in this setting.

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For that example:

$$x^2+10x+24,$$

we need two numbers whose product is $24$ and whose sum is $10$.

Factor pair of $24$	Sum
$1,24$	$25$
$2,12$	$14$
$3,8$	$11$
$4,6$	$10$

So,

$$\boxed{x^2+10x+24=(x+4)(x+6)}.$$

The order does not matter:

$$(x+4)(x+6)=(x+6)(x+4).$$

To see why the method is reliable, expand your answer:

$$\begin{aligned} (x+4)(x+6) &=x^2+6x+4x+24\\ &=x^2+10x+24. \end{aligned}$$

The expansion returns the original trinomial, so the factorization is correct.

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The sum-and-product routine

When you see

$$x^2+bx+c,$$

use this routine.

1. Read the middle coefficient $b$. This is the target sum.
2. Read the constant $c$. This is the target product.
3. List factor pairs of $c$, including appropriate negative signs.
4. Select the pair whose sum is $b$.
5. Put that pair into two binomials:

$$(x+p)(x+q).$$

Worked example: both factors positive

Factor

$$x^2+9x+18.$$

The needed pair must:

• multiply to $18$,
• add to $9$.

The positive factor pairs of $18$ are:

$$1\cdot18,\qquad 2\cdot9,\qquad 3\cdot6.$$

Their sums are $19$, $11$, and $9$. The correct pair is $3$ and $6$.

$$\boxed{x^2+9x+18=(x+3)(x+6)}$$

A quick expansion check confirms it:

$$(x+3)(x+6)=x^2+6x+3x+18=x^2+9x+18.$$

Notice that a positive constant and a positive middle coefficient led to two positive numbers. That makes sense: positive times positive is positive, and positive plus positive is positive.

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Signs: use the constant first, then the middle term

The sign of $c$, the constant term, tells you whether the two numbers have matching or opposite signs.

Signs in the trinomial	What the two numbers must look like
$c>0$	Same sign: both positive or both negative
$c<0$	Opposite signs: one positive and one negative
$c>0,\ b>0$	Both positive
$c>0,\ b<0$	Both negative
$c<0,\ b>0$	Opposite signs, with the positive number having greater magnitude
$c<0,\ b<0$	Opposite signs, with the negative number having greater magnitude

These are not separate rules to memorize blindly. They follow from integer multiplication and addition.

Worked example: both factors negative

Factor

$$x^2-8x+15.$$

We need two numbers whose product is $15$ and whose sum is $-8$.

The factor magnitudes $3$ and $5$ multiply to $15$. Because the product is positive, the signs must match. Because the sum must be negative, both numbers must be negative:

$$-3+(-5)=-8$$

and

$$(-3)(-5)=15.$$

Therefore,

$$\boxed{x^2-8x+15=(x-3)(x-5)}.$$

Check:

$$\begin{aligned} (x-3)(x-5) &=x^2-5x-3x+15\\ &=x^2-8x+15. \end{aligned}$$

Worked example: opposite signs

Factor

$$x^2+2x-63.$$

This time, the product must be $-63$. Therefore the two numbers need opposite signs.

The factor pair $7$ and $9$ has a difference of $2$. To obtain a sum of positive $2$, the larger magnitude must be positive:

$$-7+9=2$$

and

$$(-7)(9)=-63.$$

So,

$$\boxed{x^2+2x-63=(x-7)(x+9)}.$$

A useful sign check comes directly from the factorization:

$$(x-7)(x+9)$$

has a negative constant because $(-7)(+9)$ is negative. Its middle term is positive because $9-7=2$.

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Factor pairs systematically

For small constants, you may recognize the pair quickly. For larger ones, a short, organized list is safer than guessing.

Factor

$$x^2-7x-18.$$

We need a product of $-18$ and a sum of $-7$. Since the product is negative, try pairs with opposite signs:

Pair	Product	Sum
$1,-18$	$-18$	$-17$
$2,-9$	$-18$	$-7$
$3,-6$	$-18$	$-3$

The pair is $2$ and $-9$. Thus,

$$\boxed{x^2-7x-18=(x+2)(x-9)}.$$

It is tempting to put $-2$ and $9$ into the factors because they also multiply to $-18$. But their sum is $7$, not $-7$:

$$-2+9=7.$$

Their factorization would produce the wrong middle term:

$$(x-2)(x+9)=x^2+7x-18.$$

The sum condition and product condition must both be satisfied.

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Why the answer has two binomials beginning with $x$

Suppose we factor a monic trinomial:

$$x^2+bx+c.$$

The first term of each factor must be $x$, because

$$x\cdot x=x^2.$$

So the overall shape is already determined:

$$(x+\square)(x+\square).$$

Factoring consists of filling the two blanks with the right signed integers. The constants you insert must create:

• the correct constant term through multiplication;
• the correct middle term through addition.

For example, for

$$x^2-3x-28,$$

we need product $-28$ and sum $-3$. The numbers $-7$ and $4$ work:

$$(-7)(4)=-28$$

and

$$-7+4=-3.$$

Therefore,

$$\boxed{x^2-3x-28=(x-7)(x+4)}.$$

This also has a geometric interpretation. If

$$x^2-3x-28$$

represented a rectangle’s area, then the factors

$$x-7$$

and

$$x+4$$

could represent its algebraic side lengths.

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Checking and common mistakes

Expanding the two factors is the most dependable check, and you already know how to do it.

Suppose you factor

$$x^2+13x+40$$

as

$$(x+5)(x+8).$$

Verify:

$$\begin{aligned} (x+5)(x+8) &=x^2+8x+5x+40\\ &=x^2+13x+40. \end{aligned}$$

The check works.

Choosing a pair that only has the right product

For

$$x^2+11x+24,$$

both $3$ and $8$, and $4$ and $6$, multiply to $24$. But only one pair has the target sum:

$$3+8=11,$$

while

$$4+6=10.$$

Therefore,

$$x^2+11x+24=(x+3)(x+8),$$

not $(x+4)(x+6)$.

Losing signs

For

$$x^2-9x+20,$$

the constant is positive and the middle coefficient is negative. The numbers must therefore both be negative:

$$-4+(-5)=-9, \qquad (-4)(-5)=20.$$

So,

$$\boxed{x^2-9x+20=(x-4)(x-5)}.$$

Writing $(x+4)(x+5)$ would produce $+9x$, not $-9x$.

Factoring when no integer pair exists

Not every monic quadratic can be factored using integers. For example,

$$x^2+2x+6$$

would require two integers with product $6$ and sum $2$. The factor pairs of $6$ have sums $7$ and $5$, or their negative versions $-7$ and $-5$. None has sum $2$.

So this expression does not factor into binomials with integer constants. For this lesson, focus on recognizing and factoring the cases where the required integer pair does exist. Later algebra provides methods for quadratics that do not factor neatly this way.

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A compact factoring checklist

Before treating a factorization as finished, check these points:

1. The expression has the form

$$x^2+bx+c.$$

2. The two selected integers multiply to $c$.
3. The same two integers add to $b$.
4. They are placed into the factors as

$$(x+p)(x+q).$$

5. If needed, expand to verify.

For example:

$$x^2+5x-24.$$

The pair must multiply to $-24$ and add to $5$. The correct pair is $8$ and $-3$:

$$8(-3)=-24, \qquad 8+(-3)=5.$$

Thus,

$$\boxed{x^2+5x-24=(x+8)(x-3)}.$$

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Key takeaways

Factoring a monic quadratic trinomial reverses the expansion pattern from the previous lesson:

$$(x+p)(x+q)=x^2+(p+q)x+pq.$$

To factor

$$x^2+bx+c,$$

find integers that add to $b$ and multiply to $c$, then place them in

$$(x+p)(x+q).$$

Use the sign of $c$ to determine whether the pair has matching or opposite signs, and use the sign of $b$ to decide which signs give the required sum. Finally, expand the factors if you want a reliable verification.

Next, you will use factored expressions to solve quadratic equations. A factorization such as

$$(x-3)(x+5)=0$$

will let you identify the two input values that make the quadratic equal zero—exactly the $x$-intercepts that matter for graphing parabolas.