Factoring Quadratic Trinomials of the Form x² + bx + c

Hello again. Last lesson, you learned that factoring — פירוק לגורמים — reverses distribution: you take a shared factor outside parentheses. Now we use the same reverse-thinking idea for a special three-term expression.

This lesson is about factoring a quadratic trinomial — תלת־איבר ריבועי — of the form

$$x^2+bx+c$$

By the end, you will be able to turn it into two binomials — דו־איברים — such as

$$(x+2)(x+3)$$

You will also know how the plus and minus signs tell you which numbers to use. This is important preparation for the next lesson, where these factored expressions become quadratic equations that you can solve.

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The pattern: reverse expansion

A trinomial has three terms — three parts separated by $+$ or $-$. In this lesson, its first term is always $x^2$, with no number in front of it except the hidden $1$:

$$x^2+bx+c$$

Here is the key expansion pattern:

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

When we distribute:

$$(x+p)(x+q)=x^2+qx+px+pq$$

Combine the two middle $x$-terms:

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

Compare this with:

$$x^2+bx+c$$

This shows the two facts you need:

$$p+q=b$$ $$pq=c$$

In words:

• Find two numbers that add to the middle coefficient $b$.
• The same two numbers must multiply to the last number $c$.

Then write:

$$x^2+bx+c=(x+p)(x+q)$$

Essential vocabulary

English	Hebrew	Meaning
quadratic	ריבועי	An expression whose highest power is $2$
trinomial	תלת־איבר	An expression with three terms
binomial	דו־איבר	An expression with two terms, often in parentheses
factor	גורם	A quantity being multiplied
factor pair	זוג גורמים	Two numbers whose product is a given number
coefficient	מקדם	Number multiplying a variable; $b$ is the coefficient of $x$
constant term	האיבר החופשי	A number with no variable; $c$ is the constant
expand / distribute	לפתוח סוגריים / חוק הפילוג	Multiply out parentheses
factor	לפרק לגורמים	Rewrite an expression as multiplication

A useful test-reading note: if the question says “Factor completely” — פרק לגורמים באופן מלא — first check whether there is a GCF. For expressions of the exact form $x^2+bx+c$, there is usually no GCF, so you can use the method in this lesson directly.

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A reliable factoring routine

Suppose you need to factor:

$$x^2+7x+12$$

Step 1: Set up the parentheses

Because:

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

the answer must begin with:

$$(x\square )(x\square )$$

The boxes will contain numbers and signs.

Step 2: Identify $b$ and $c$

For

$$x^2+7x+12$$

we have:

$$b=7$$ $$c=12$$

So find two numbers that:

• multiply to $12$;
• add to $7$.

List positive factor pairs of $12$:

Factor pair	Sum
$1,12$	$13$
$2,6$	$8$
$3,4$	$7$

The pair $3$ and $4$ works because:

$$3\cdot 4=12$$

and

$$3+4=7$$

Step 3: Put those numbers into the binomials

$$x^2+7x+12=(x+3)(x+4)$$

The order does not matter:

$$(x+4)(x+3)$$

is exactly the same answer.

Step 4: Check by expanding

On a written test, a quick expansion check catches sign errors:

$$(x+3)(x+4)=x^2+4x+3x+12$$ $$=x^2+7x+12$$

It matches the original trinomial, so the factorization is correct.

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Signs are clues, not decoration

The signs of $b$ and $c$ tell you what signs $p$ and $q$ must have.

Remember the two conditions:

$$p+q=b$$ $$pq=c$$

The product $pq$ tells you whether the signs are the same or different.

Sign of $c$	What this means for $p$ and $q$
$c>0$	Same signs: both positive or both negative
$c<0$	Different signs: one positive and one negative

Then the sum $p+q=b$ tells you which sign arrangement is correct.

When $c$ is positive

A positive product comes from two positive numbers or two negative numbers.

• If $b$ is positive, use two positive numbers.
• If $b$ is negative, use two negative numbers.

For example:

$$x^2-8x+12$$

The numbers must multiply to $12$ and add to $-8$.

Since the product is positive, the signs match. Since the sum is negative, both numbers are negative:

$$-2\cdot (-6)=12$$ $$-2+(-6)=-8$$

Therefore:

$$x^2-8x+12=(x-2)(x-6)$$

Check the middle terms:

$$(x-2)(x-6)=x^2-6x-2x+12$$ $$=x^2-8x+12$$

When $c$ is negative

A negative product needs opposite signs: one number positive, one number negative.

The number with the larger absolute value decides the sign of the sum.

Consider:

$$x^2+2x-15$$

We need two numbers that multiply to $-15$ and add to $2$.

The factor pairs of $15$ are:

$$1\text{ and }15$$ $$3\text{ and }5$$

Because the product must be negative, try opposite signs. The pair $5$ and $3$ has a difference of $2$. To get a positive sum, the $5$ must be positive:

$$5+(-3)=2$$ $$5\cdot (-3)=-15$$

So:

$$x^2+2x-15=(x+5)(x-3)$$

Check it:

$$(x+5)(x-3)=x^2-3x+5x-15$$ $$=x^2+2x-15$$

The image’s wording “make $2$ by addition or subtraction” means exactly this signed-number check:

$$5+(-3)=2$$

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The four sign situations

You do not need to memorize a complicated rule. Still, this compact chart can help under test pressure.

Trinomial pattern	Signs inside the factors	Example result
$x^2+\text{positive }x+\text{positive}$	both $+$	$(x+p)(x+q)$
$x^2-\text{positive }x+\text{positive}$	both $-$	$(x-p)(x-q)$
$x^2+\text{positive }x-\text{positive}$	different; larger number is $+$	$(x-p)(x+q)$
$x^2-\text{positive }x-\text{positive}$	different; larger number is $-$	$(x-p)(x+q)$

The last two rows may look alike in writing, so rely on the add check, not only the chart.

For example, factor:

$$x^2-x-12$$

We need a product of $-12$ and a sum of $-1$. The factor pair $3$ and $4$ is one apart. The larger number must be negative because the sum must be negative:

$$3+(-4)=-1$$ $$3\cdot (-4)=-12$$

Therefore:

$$x^2-x-12=(x+3)(x-4)$$

A common error would be:

$$(x-3)(x+4)$$

But its middle coefficient is positive:

$$-3+4=1$$

So it would expand to $x^2+x-12$, not the original expression.

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A fast scratch-paper method

For each question, make a tiny two-column note:

$$\begin{array}{cc}\text{multiply}& \text{add}\\ c& b\end{array}$$

For example:

$$x^2-5x-14$$

Write:

$$\begin{array}{cc}\text{multiply}& \text{add}\\ -14& -5\end{array}$$

The factors of $14$ are $1,14$ and $2,7$. You need opposite signs because the product is negative. The pair $2$ and $7$ has difference $5$, and the larger number must be negative:

$$2+(-7)=-5$$

So:

$$x^2-5x-14=(x+2)(x-7)$$

This written routine is safer than guessing:

1. Check for a GCF first.
2. Write down $b$, the middle coefficient.
3. Write down $c$, the constant term.
4. List factor pairs of $|c|$.
5. Assign signs so the product is $c$.
6. Check that the signed numbers add to $b$.
7. Write the two binomials.
8. Expand briefly to verify if you have time.

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When no integer pair works

Some trinomials do not factor into integer binomials. For example:

$$x^2+7x-6$$

We would need two integers with product $-6$ and sum $7$.

Possible opposite-sign pairs are based on $1,6$ and $2,3$:

$$6+(-1)=5$$ $$3+(-2)=1$$

Switching the signs gives negative sums, not $7$. No integer pair works.

For this course, call such a trinomial prime — ראשוני — or not factorable over the integers. Do not invent a pair just because the numbers look close. The two checks, product and sum, must both work.

Later mathematics has other ways to solve or rewrite these quadratics, but the next lesson will focus only on equations that do factor cleanly.

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Key takeaways

To factor a trinomial of the form

$$x^2+bx+c$$

find two integers $p$ and $q$ such that:

$$p+q=b$$

and

$$pq=c$$

Then write:

$$x^2+bx+c=(x+p)(x+q)$$

Keep these sign facts in mind:

• Positive $c$: the two numbers have the same sign.
• Negative $c$: the two numbers have different signs.
• The middle coefficient $b$ confirms whether the final sum must be positive or negative.
• Always verify by expanding if a sign feels uncertain.

Next, you will use factored quadratics to solve equations such as

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

using the zero-product property — חוק המכפלה האפסית.