8th Grade Common Core Math Worksheets - Free Printable

Let’s solve each problem one by one, step by step. We’ll use the rules of exponents:

- When multiplying powers with the same base, add the exponents:
\( x^a \cdot x^b = x^{a+b} \)
- Multiply coefficients (numbers) normally.
- Handle negative signs carefully.

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Problem 1: Simplify \( x^2 \cdot 4x^3 \cdot y^4 \cdot 4y^2 \)

Step 1: Group numbers and like variables.

Numbers: \( 4 \cdot 4 = 16 \)

\( x \) terms: \( x^2 \cdot x^3 = x^{2+3} = x^5 \)

\( y \) terms: \( y^4 \cdot y^2 = y^{4+2} = y^6 \)

So total: \( 16x^5y^6 \)

✔ Final Answer for #1: [A] 16x⁵y⁶

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Problem 2: Simplify \( (2xy^2)(5x^4y^4) \)

Multiply coefficients: \( 2 \cdot 5 = 10 \)

\( x \cdot x^4 = x^{1+4} = x^5 \)

\( y^2 \cdot y^4 = y^{2+4} = y^6 \)

Result: \( 10x^5y^6 \)

✔ Final Answer for #2: [C] 10x⁵y⁶

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Problem 3: Simplify \( (7xy^2)(9xy^4) \)

Coefficients: \( 7 \cdot 9 = 63 \)

\( x \cdot x = x^{1+1} = x^2 \)

\( y^2 \cdot y^4 = y^{2+4} = y^6 \)

Result: \( 63x^2y^6 \)

✔ Final Answer for #3: [B] 63x²y⁶

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Problem 4: Simplify \( (-9x^3y^4)(-6x^3y^2) \)

Coefficients: \( -9 \cdot -6 = +54 \) (negative times negative is positive)

\( x^3 \cdot x^3 = x^{3+3} = x^6 \)

\( y^4 \cdot y^2 = y^{4+2} = y^6 \)

Result: \( 54x^6y^6 \)

✔ Final Answer for #4: [D] 54x⁶y⁶

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Problem 5: Simplify \( (3x^3y)(-4xy^2) \)

Coefficients: \( 3 \cdot -4 = -12 \)

\( x^3 \cdot x = x^{3+1} = x^4 \)

\( y \cdot y^2 = y^{1+2} = y^3 \)

Result: \( -12x^4y^3 \)

✔ Final Answer for #5: [B] -12x⁴y³

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Problem 6: Simplify \( (-6x^2y^4)(-8x^2y) \)

Coefficients: \( -6 \cdot -8 = +48 \)

\( x^2 \cdot x^2 = x^{2+2} = x^4 \)

\( y^4 \cdot y = y^{4+1} = y^5 \)

Result: \( 48x^4y^5 \)

✔ Final Answer for #6: [C] 48x⁴y⁵

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Problem 7: Simplify \( (-5x^4y^3)(2x^3y^3) \)

Coefficients: \( -5 \cdot 2 = -10 \)

\( x^4 \cdot x^3 = x^{4+3} = x^7 \)

\( y^3 \cdot y^3 = y^{3+3} = y^6 \)

Result: \( -10x^7y^6 \)

✔ Final Answer for #7: [B] -10x⁷y⁶

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Problem 8: Show two ways to use a calculator to find the value of \( 6^2 \cdot 6^3 \)

Way 1: Calculate each power separately, then multiply.

→ Type: `6 ^ 2` → get 36
→ Type: `6 ^ 3` → get 216
→ Then: `36 * 216` → get 7776

Way 2: Use exponent rule first: \( 6^2 \cdot 6^3 = 6^{2+3} = 6^5 \)

→ Type: `6 ^ 5` → get 7776

Both give same answer!

✔ Final Answer for #8:
Two ways:
① Compute 6² = 36, 6³ = 216, then 36 × 216 = 7776
② Add exponents: 6^(2+3) = 6⁵ = 7776

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Problem 9: Write \( 2x^9 \) as the product of two powers with the same base.

We can split the exponent 9 into any two numbers that add to 9.

Example: \( x^9 = x^4 \cdot x^5 \), so:

\( 2x^9 = 2 \cdot x^4 \cdot x^5 \)

But we want “product of two powers” — probably meaning two terms that are both powers (including coefficient?).

Actually, since 2 is not a power of x, maybe they mean write it as product of two expressions that are powers? But 2 is constant.

Wait — perhaps they allow splitting the variable part only.

Better interpretation: Write as product of two powers of x, multiplied by 2? But question says “as the product of two powers with the same base”.

Hmm — maybe they consider 2 as \( 2^1 \), but base must be same.

Alternative: Maybe they want something like \( (2x^a)(x^b) \) where a + b = 9? But then bases aren’t same unless we include 2 in base — which doesn’t make sense.

Wait — perhaps misread. Let me re-read: “Write \( 2x^9 \) as the product of two powers with the same base.”

This is tricky because 2 is not a power of x. Unless...

Maybe they mean: express as \( (something)^m \cdot (something)^n \) with same base.

One possible way: \( 2x^9 = (2x^3)(x^6) \) — but again, different bases if you think of 2x³ vs x⁶.

Another idea: Perhaps factor out constants differently? Not really.

Wait — maybe they accept: \( 2x^9 = x^4 \cdot 2x^5 \) — still not same base.

I think there might be a mistake in interpretation.

Perhaps the question allows the coefficient to stay outside? Or maybe it's expecting us to write as \( (x^a)(2x^b) \) but that’s not two powers with same base.

Let me check standard approach.

Actually, in some contexts, “powers with the same base” refers only to the variable part. So perhaps:

\( 2x^9 = 2 \cdot x^4 \cdot x^5 \), and say the two powers are \( x^4 \) and \( x^5 \), but then 2 is extra.

The question says “write 2x⁹ as the product of two powers”, implying the entire expression equals product of two terms, each being a power (with same base).

That would require writing 2 as a power too — e.g., if base is 2, but then x is different.

Unless... maybe they mean base x, and treat 2 as \( 2x^0 \)? But that’s stretching.

Perhaps it's a typo or poor wording. Common version of this question is to write \( x^9 \) as product of two powers — then easy: \( x^4 \cdot x^5 \).

Given that, I’ll assume they meant to ignore the 2 or include it in one term.

Best practical answer: \( 2x^9 = (2x^4)(x^5) \) — but bases are not same.

Wait — another thought: Maybe “powers” here means monomials, and “same base” refers to variable base.

In many textbooks, they’d accept: \( 2x^9 = x^3 \cdot 2x^6 \), but again, not same base.

I think the intended answer is to split the exponent on x, keeping 2 attached to one part.

But let’s look at Problem 10 for clue.

Problem 10: Write \( 8x^8 \) as product of two powers with same base.

Similarly, 8 is 2³, but x is different.

Perhaps for Problem 9: \( 2x^9 = (2x^3)(x^6) \) — no.

Wait — what if we write 2 as \( (\sqrt{2})^2 \)? Too complicated.

I recall sometimes they do: \( 2x^9 = (x^a)(2x^b) \) and call it done, even though technically bases differ.

But to satisfy “same base”, perhaps only the x part matters.

Let me propose: Since 2 is constant, we can write:

\( 2x^9 = 2 \cdot x^4 \cdot x^5 \), and say the two powers are \( x^4 \) and \( x^5 \), but then 2 is multiplier.

The question says “as the product of two powers”, so likely expects two factors.

Perhaps: \( (2x^4) \cdot (x^5) \), and consider “base” as x for the variable part.

I think for school level, they expect:

For #9: \( 2x^9 = (2x^4)(x^5) \) or \( (x^3)(2x^6) \), etc.

But to have “same base”, maybe they want both terms to have base x, so 2 must be incorporated.

Another idea: Write 2 as \( 2^1 \), but then base is 2, not x.

Unless the base is not specified — but it says “with the same base”, implying we choose a base.

Perhaps for #9, we can write: \( 2x^9 = (2^{1/2} x^{9/2})^2 \) — no, too advanced.

I think there's a better way.

Let me search my memory: In some curricula, they allow writing as \( k \cdot x^a \cdot x^b \), and call x^a and x^b the two powers.

But the expression is 2x^9, which is already a single term.

Perhaps the question means: express it as a product of two expressions, each of which is a power function with the same base variable.

So, for example: \( 2x^9 = (x^4) \cdot (2x^5) \)

And we say the base is x for both, even though one has a coefficient.

I think that's acceptable for middle school.

To be safe, I'll provide a common answer.

For #9: One possible answer is \( (2x^3)(x^6) \)

But let's see what makes sense with #10.

Problem 10: Write \( 8x^8 \) as product of two powers with same base.

8 is 2^3, but again.

Note that 8x^8 = (2x^2)^3? No, (2x^2)^3 = 8x^6.

8x^8 = (2x^4)^2? (2x^4)^2 = 4x^8 — not 8.

8x^8 = (√8 x^4)^2 — messy.

Better: 8x^8 = 8 * x^8 = (8x^4)(x^4) — then both have x^4, but different coefficients.

Or 8x^8 = (2x^4)(4x^4) — now both have x^4, and 2*4=8.

Oh! That works.

Similarly for #9: 2x^9 = (2x^4)(x^5) — but exponents different.

To have same exponent? Not necessary.

The question doesn't say same exponent, just same base.

So for #9: We can write 2x^9 = (x^a)(2x^b) with a+b=9.

But to make it "two powers", perhaps they want both to be pure powers, but 2 is there.

Another thought: Perhaps "powers" includes the coefficient if it's a perfect power, but 2 is not.

I recall that in some problems, they do: for 2x^9, write as x^4 * 2x^5, and that's fine.

For consistency, let's do for #9: \( 2x^9 = (2x^4)(x^5) \)

But let's calculate: (2x^4)(x^5) = 2x^{9} — yes.

And both have base x.

Similarly for #10: 8x^8 = (2x^4)(4x^4) = 8x^8, and both have base x.

Or (8x^4)(x^4), etc.

So I'll go with that.

For #9: \( 2x^9 = (2x^4)(x^5) \) or any split.

To make it neat, perhaps use equal exponents if possible, but 9 is odd.

Common choice: \( x^4 \cdot 2x^5 \)

But order doesn't matter.

I think [A] isn't given, so we need to write an expression.

Since it's open-ended, I'll provide one.

Similarly for #10.

So for #9: One way is \( (x^3)(2x^6) \)

But let's pick \( (2x^4)(x^5) \)

Actually, to avoid confusion, perhaps write as \( 2 \cdot x^4 \cdot x^5 \), but that's three factors.

The question says "product of two powers", so two factors.

So I'll say: \( (2x^4) \times (x^5) \)

But is 2x^4 considered a "power"? Technically, a power is like x^n, but in context, it might be accepted.

Perhaps they mean write as (constant * x^a) * (x^b), and call it two powers with base x.

I think for the purpose of this worksheet, that's fine.

So for #9: \( 2x^9 = (2x^4)(x^5) \)

For #10: \( 8x^8 = (2x^4)(4x^4) \) or (8x^4)(x^4), etc.

Let's choose symmetric for #10: since 8=2*4, and 8/2=4, so (2x^4)(4x^4)

Yes.

So:

✔ Final Answer for #9: \( (2x^4)(x^5) \) or equivalent

✔ Final Answer for #10: \( (2x^4)(4x^4) \) or equivalent

But to box them, since it's not multiple choice, I'll write the expressions.

Now, compiling all final answers.

──────────────────────────────────────

Final Answer:
1. [A] 16x⁵y⁶
2. [C] 10x⁵y⁶
3. [B] 63x²y⁶
4. [D] 54x⁶y⁶
5. [B] -12x⁴y³
6. [C] 48x⁴y⁵
7. [B] -10x⁷y⁶
8. Two ways:
- Calculate 6² = 36, 6³ = 216, then 36 × 216 = 7776
- Use exponent rule: 6² · 6³ = 6⁵ = 7776
9. Example: \( (2x^4)(x^5) \)
10. Example: \( (2x^4)(4x^4) \)