Solve (5-x)(6+2x)=-2+4x Using the Distributive Property

Q: Why do I get two answers instead of just one?

This is a quadratic equation (has $ x^2 $), so it can have up to 2 solutions! When you get $ x^2 = 16 $, both +4 and -4 work because $ (+4)^2 = 16 $ and $ (-4)^2 = 16 $.

Q: How do I keep track of all the terms when distributing?

Use the FOIL method or make a systematic list: First terms, Outer terms, Inner terms, Last terms. Write each step clearly: $ 5 \cdot 6, 5 \cdot 2x, -x \cdot 6, -x \cdot 2x $.

Q: How do I know when to use the distributive property?

Use it whenever you see parentheses next to each other like $ (a+b)(c+d) $. You must distribute before you can combine like terms or solve the equation.

Q: Can I check both answers the same way?

Absolutely! Substitute $ x = 4 $ and $ x = -4 $ separately into the original equation $ (5-x)(6+2x) = -2+4x $. Both should make the equation true.

Quadratic Equations with Distributive Property

Solve the following equation using the distributive property:

$(5-x)(6+2x)=-2+4x$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Open parentheses properly, multiply each factor by each factor

00:23 Calculate the products

00:44 Collect terms

01:18 Isolate X

01:28 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following equation using the distributive property:

$(5-x)(6+2x)=-2+4x$

Step-by-step solution

To solve the given equation $(5-x)(6+2x)=-2+4x$ , we'll apply the following steps to use the distributive property:

Step 1: Expand the left side using the distributive property:
$(5-x)(6+2x) = 5 \cdot 6 + 5 \cdot 2x - x \cdot 6 - x \cdot 2x$ $= 30 + 10x - 6x - 2x^2$ Finally, combine like terms:
$= -2x^2 + 4x + 30$
Step 2: Set the equation against the right side:
$-2x^2 + 4x + 30 = -2 + 4x$
Step 3: Move all terms to one side of the equation:
$-2x^2 + 4x + 30 - 4x - (-2) = 0$ Simplify to:
$-2x^2 + 32 = 0$
Step 4: Solve the quadratic equation:
Factor out the common term:
$-2(x^2 - 16) = 0$
$x^2 - 16 = 0$
Step 5: Solve for $x$ :
$x^2 = 16$
$x = \pm 4$

Therefore, the solution to the equation is $x = \pm 4$ .

Final Answer

$x=±4$

Key Points to Remember

Essential concepts to master this topic

Distribution: Multiply each term by each term systematically
Technique: $(5-x)(6+2x) = 30 + 10x - 6x - 2x^2$
Check: Substitute both solutions back into original equation ✓

Common Mistakes

Avoid these frequent errors

Forgetting the negative sign when distributing
Don't ignore negative signs when multiplying = wrong coefficients! Many students write +6x instead of -6x when distributing -x times 6. Always track negative signs carefully through each multiplication step.

Practice Quiz

Test your knowledge with interactive questions

It is possible to use the distributive property to simplify the expression below?

What is its simplified form?

$$ (ab)(c d) $$

$$

FAQ

Everything you need to know about this question

Why do I get two answers instead of just one?

This is a quadratic equation (has $x^2$ ), so it can have up to 2 solutions! When you get $x^2 = 16$ , both +4 and -4 work because $(+4)^2 = 16$ and $(-4)^2 = 16$ .

How do I keep track of all the terms when distributing?

Use the FOIL method or make a systematic list: First terms, Outer terms, Inner terms, Last terms. Write each step clearly: $5 \cdot 6, 5 \cdot 2x, -x \cdot 6, -x \cdot 2x$ .

What if I get a different quadratic form?

No problem! The key is getting everything on one side equal to zero. Whether you get $-2x^2 + 32 = 0$ or $2x^2 - 32 = 0$ , you'll get the same final answers.

How do I know when to use the distributive property?

Use it whenever you see parentheses next to each other like $(a+b)(c+d)$ . You must distribute before you can combine like terms or solve the equation.

Can I check both answers the same way?

Absolutely! Substitute $x = 4$ and $x = -4$ separately into the original equation $(5-x)(6+2x) = -2+4x$ . Both should make the equation true.

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