Q. Use the distributive property to expand \( -3(8x-4)\).

Q: How do I apply the distributive property to -3(8x-4) ?.

Multiply -3 by each term: -3\cdot8x=-24x and -3\cdot(-4)=12. So -3(8x-4)=-24x+12.

Q: Why is a negative times a negative positive?

Because sign rules give (-a)(-b)=ab . Multiplying two negatives reverses the sign twice, producing a positive result.

Q: How can I check my expansion is correct?.

Factor the result back: -24x+12=-3(8x-4). Or plug a test value (e.g., x=1): left -3(8-4)=-12, right -24+12=-12..

Q: How do I factor the expanded result back into the original form?

Factor out the common factor -3: -24x+12=-3(8x-4). Divide coefficients by -3 to recover 8x-4.

Q: How do I expand when the outside coefficient is a fraction or decimal?

Same rule: multiply it into each term. Example: \tfrac{-1}{2}(8x-4)=-4x+2. For decimals, -0.5(8x-4)=-4x+2..

Q: How does this extend to more terms or variables inside parentheses?

Distribute to every term. Example: -3(2x+3y-5)=-6x-9y+15; multiply -3 by each of 2x, 3y, and -5..

Answer

Multiply −3 by each term: \( -3(8x-4)=(-3)(8x)+(-3)(-4)=-24x+12.\)

Final result: \( -24x+12.\)

Detailed Explanation

Write the original expression: \( -3(8x – 4) \).
Recall the distributive property: \( a(b+c)=ab+ac \). Because subtraction is addition of a negative, rewrite the inside as \(8x+(-4)\). Thus \( -3(8x – 4)= -3(8x+(-4))\).
Apply the distributive property by multiplying \(-3\) by each term: \( -3(8x+(-4)) = -3\cdot 8x + -3\cdot(-4)\).
Compute each product: \( -3\cdot 8x = -24x\) and \( -3\cdot(-4)=12\).
Combine the results to get the expanded form: \( -24x + 12\).

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Algebra FAQs

How do I apply the distributive property to \( -3(8x-4) \)?.

Multiply \(-3\) by each term: \(-3\cdot8x=-24x\) and \(-3\cdot(-4)=12\). So \(-3(8x-4)=-24x+12\).

Why is a negative times a negative positive?

Because sign rules give \( (-a)(-b)=ab \). Multiplying two negatives reverses the sign twice, producing a positive result.

How can I check my expansion is correct?.

Factor the result back: \(-24x+12=-3(8x-4)\). Or plug a test value (e.g., \(x=1\)): left \(-3(8-4)=-12\), right \(-24+12=-12\)..

How do I factor the expanded result back into the original form?

Factor out the common factor \(-3\): \(-24x+12=-3(8x-4)\). Divide coefficients by \(-3\) to recover \(8x-4\).

What are common mistakes students make here?

Forgetting to distribute the negative, dropping a negative sign, or mis-multiplying coefficients. Always multiply the outer coefficient by every term inside the parentheses.

Does the distributive property work for addition inside parentheses too?

How do I expand when the outside coefficient is a fraction or decimal?

Same rule: multiply it into each term. Example: \(\tfrac{-1}{2}(8x-4)=-4x+2\). For decimals, \(-0.5(8x-4)=-4x+2\)..

After expanding, how do I combine like terms?

Group terms with the same variable and add their coefficients (e.g., \(-24x+5x=-19x\)). Combine constants separately (e.g., \(12+3=15\)).

How does this extend to more terms or variables inside parentheses?

Distribute to every term. Example: \(-3(2x+3y-5)=-6x-9y+15\); multiply \(-3\) by each of \(2x\), \(3y\), and \(-5\)..

Expand -3(8x-4) via distributive law.
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