Solving and Verifying Factorable Quadratic Equations

Hi again. In the previous lesson, you learned to factor trinomials such as

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

Factoring turns a quadratic expression — ביטוי ריבועי — into multiplication. Today you will use that multiplication to solve a quadratic equation — משוואה ריבועית.

By the end of this lesson, you should be able to factor a quadratic equation, use the zero-product property — חוק המכפלה האפסית — to find all its solutions, and check each solution by substitution. This is a high-priority test skill: the key is to get the equation into a factored form with $0$ on one side.

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Why a product equal to zero is special

Consider this equation:

$$(x-7)(x+2)=0$$

The left side is a product: two factors — גורמים — multiplied together.

The only way multiplication can give $0$ is for at least one factor to be $0$. For example:

$$12\cdot 0=0$$ $$(-4)\cdot 0=0$$

But two nonzero numbers cannot multiply to make zero:

$$3\cdot 5=15$$ $$(-2)\cdot 7=-14$$

This fact is called the zero-product property:

$$\text{If }A\cdot B=0,\text{ then }A=0\text{ or }B=0$$

The word or — או — matters. Usually, one factor is zero for one solution, and the other factor is zero for a different solution. They do not both need to be zero at the same time.

Here is the same idea visually:

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Solving an equation already in factored form

Let us solve:

$$(x-7)(x+2)=0$$

Step 1: Set each factor equal to zero

$$x-7=0$$

or

$$x+2=0$$

Step 2: Solve both linear equations

For the first factor, add $7$ to both sides:

$$x=7$$

For the second factor, subtract $2$ from both sides:

$$x=-2$$

Therefore, the solutions — פתרונות — are:

$$x=7\text{ or }x=-2$$

A useful shortcut is to notice that each factor gives the opposite signed number:

Factor	Value that makes it zero
$x-7$	$x=7$
$x+2$	$x=-2$
$x+5$	$x=-5$
$x-11$	$x=11$

Do not just copy the signs inside the parentheses. For instance, $x+2=0$ does not give $x=2$. Since you subtract $2$ from both sides, it gives:

$$x=-2$$

Verify by substitution — הצבה

To verify — לוודא / לבדוק — a solution means substitute it back into the original equation and check that both sides are equal.

Check $x=7$:

$$(7-7)(7+2)=0$$ $$0\cdot 9=0$$

So $x=7$ works.

Check $x=-2$:

$$(-2-7)(-2+2)=0$$ $$(-9)\cdot 0=0$$

So $x=-2$ also works.

Notice that when one factor becomes zero, you do not need the other factor to be zero. Any number multiplied by zero is zero.

Vocabulary for test questions

English	Hebrew	Meaning
solve	פתור	Find the value or values of the variable
solution	פתרון	A value that makes the equation true
root	שורש	Another name for a solution of a polynomial equation
zero	אפס	A value of $x$ that makes an expression equal $0$
verify / check	בדוק / ודא	Substitute your answer to prove it works
substitute	הצב	Replace the variable with a number
zero-product property	חוק המכפלה האפסית	If a product is $0$, at least one factor is $0$
factor	גורם	A part of a multiplication expression

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Factors that contain a number in front of $x$

Sometimes a factor is not just $x+number$. You may see:

$$(2x-3)(3x-5)=0$$

The zero-product step stays exactly the same:

$$2x-3=0$$

or

$$3x-5=0$$

Now solve each linear equation carefully.

For the first factor:

$$2x-3=0$$

Add $3$ to both sides:

$$2x=3$$

Divide both sides by $2$:

$$x=\frac{3}{2}$$

For the second factor:

$$3x-5=0$$

Add $5$ to both sides:

$$3x=5$$

Divide both sides by $3$:

$$x=\frac{5}{3}$$

So the complete answer is:

$$x=\frac{3}{2}\text{ or }x=\frac{5}{3}$$

The quadratic equation can have fractions as solutions. That is completely normal.

A quick verification

Check $x=\frac{3}{2}$ in the original equation:

$$\left(2\left(\frac{3}{2}\right)-3\right)\left(3\left(\frac{3}{2}\right)-5\right)=0$$

The first factor becomes:

$$2\left(\frac{3}{2}\right)-3=3-3=0$$

Therefore the whole product is $0$, so the value works.

You can similarly check $x=\frac{5}{3}$: this time the second factor becomes zero.

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From a quadratic equation to a factored equation

Often the test will not give you the factors immediately. It may give you an equation like:

$$x^2-3x-10=0$$

This is where the factoring skill from the previous lesson comes in.

You need two numbers that:

• multiply to $-10$;
• add to $-3$.

Those numbers are $2$ and $-5$:

$$2\cdot (-5)=-10$$ $$2+(-5)=-3$$

So factor first:

$$x^2-3x-10=0$$ $$(x+2)(x-5)=0$$

Only after factoring may you use the zero-product property:

$$x+2=0$$

or

$$x-5=0$$

Solve:

$$x=-2$$ $$x=5$$

Therefore:

$$\boxed{x=-2\text{ or }x=5}$$

Verify both answers in the original equation

Check $x=-2$:

$$(-2{)}^2-3(-2)-10$$ $$4+6-10=0$$

Check $x=5$:

$$5^2-3(5)-10$$ $$25-15-10=0$$

Both values make the original equation equal to $0$, so both are correct.

A test answer should normally list both solutions unless the two solutions happen to be the same.

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A special but common case: a GCF gives one solution of zero

Recall that a greatest common factor (GCF) — גורם משותף גדול ביותר — is a factor shared by every term.

Solve:

$$x^2+5x=0$$

Factor out the shared $x$:

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

Now set each factor equal to zero:

$$x=0$$

or

$$x+5=0$$

The second equation gives:

$$x=-5$$

So:

$$\boxed{x=0\text{ or }x=-5}$$

This is an important pattern. If every term in a quadratic has an $x$, one solution is often $x=0$.

Check:

$$0^2+5(0)=0$$ $$(-5{)}^2+5(-5)=25-25=0$$

Both work.

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The equation must equal zero first

The zero-product property works only when the product equals $0$.

For example, this is ready to solve:

$$(x+3)(x-4)=0$$

But this is not ready:

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

You cannot set $x+3=0$ or $x-4=0$ there, because a product equaling $12$ does not require either factor to be zero.

Sometimes you must rearrange the equation before factoring. Solve:

$$x^2+2x=15$$

First, subtract $15$ from both sides so that the right side is zero:

$$x^2+2x-15=0$$

Now factor. Find two numbers that multiply to $-15$ and add to $2$:

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

Therefore:

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

Set each factor equal to zero:

$$x+5=0$$

or

$$x-3=0$$

So:

$$x=-5$$ $$x=3$$

Check in the original equation $x^2+2x=15$.

For $x=-5$:

$$(-5{)}^2+2(-5)=25-10=15$$

For $x=3$:

$$3^2+2(3)=9+6=15$$

Both solutions are valid.

The complete test routine is:

1. Put all terms on one side, with $0$ on the other side.
2. Factor completely.
3. Set each factor equal to $0$.
4. Solve each resulting linear equation.
5. Substitute both answers into the original equation to verify.

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One repeated solution

Occasionally, the same factor appears twice:

$$(x-4{)}^2=0$$

This means:

$$(x-4)(x-4)=0$$

Both factors produce the same equation:

$$x-4=0$$

Thus:

$$x=4$$

There is only one distinct solution. You may hear that $x=4$ is a double root — שורש כפול — because the factor is repeated, but for now the important point is simple: write the answer once.

Verification:

$$(4-4{)}^2=0^2=0$$

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Common mistakes to avoid

1. Using the zero-product property before factoring

This is incorrect:

$$x^2-3x-10=0$$ $$x^2=0\text{ or }-3x=0\text{ or }-10=0$$

The left side is a sum, not a product. Factor it first.

2. Forgetting that a plus in a factor gives a negative solution

$$x+8=0$$ $$x=-8$$

The solution has the opposite sign.

3. Factoring correctly but solving only one factor

From:

$$(x+2)(x-5)=0$$

you need both:

$$x+2=0$$

and

$$x-5=0$$

A quadratic commonly has two solutions.

4. Not making one side equal to zero

The equation

$$x^2+2x=15$$

must become:

$$x^2+2x-15=0$$

before factoring and applying the zero-product property.

5. Skipping the check

A sign error in factoring can create a wrong answer. A quick substitution tells you whether the original equation is truly satisfied.

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Key takeaways

To solve a factorable quadratic equation:

$$\text{quadratic expression}=0$$

first write it as:

$$(\text{factor})(\text{factor})=0$$

Then apply the zero-product property:

$$A\cdot B=0$$

means:

$$A=0\text{ or }B=0$$

Solve each linear equation and verify each result by substitution into the original equation.

Remember these essentials:

• You can use the zero-product property only when one side is $0$.
• Factor before splitting into two equations.
• A factor $x+a$ produces $x=-a$.
• A factor $x-a$ produces $x=a$.
• Check both proposed solutions.

Next, you will begin algebraic fractions — שברים אלגבריים — by learning how to identify values that are not allowed because they make a denominator equal to zero.