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

$(3+20)\times(12+4)=$

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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