Hard GCSE Index laws in algebra Questions

Challenging, exam-style GCSE Index laws in algebra questions with worked solutions. Stretch yourself on the hardest multiplication law, division law, coefficients, power of a product problems.

multiplication lawdivision lawcoefficientspower of a productnegative indicesfractional indices
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Simplify (2x3)2×(3x2y)212x5y\frac{(2x^3)^2 \times (3x^2 y)^2}{12x^5 y}.
Show worked solution

Worked solution

  1. Deal with the first bracket

    (2x3)2(2x^3)^2

    Apply the power 2 to the first bracket.

  2. Square its coefficient

    22=42^2 = 4

    Two squared is four.

  3. Multiply its indices

    x3×2=x6x^{3 \times 2} = x^{6}

    Three times two is six, so the first bracket is 4x to the power 6.

  4. Deal with the second bracket

    (3x2y)2(3x^2 y)^2

    Apply the power 2 to the second bracket.

  5. Square its coefficient

    32=93^2 = 9

    Three squared is nine.

  6. Multiply its x indices

    x2×2=x4x^{2 \times 2} = x^{4}

    Two times two is four.

  7. Multiply its y indices

    y1×2=y2y^{1 \times 2} = y^{2}

    The y has index 1, so 1 times 2 is 2, giving 9x to the power 4 y squared.

  8. Multiply the two brackets

    4x6×9x4y24x^{6} \times 9x^{4}y^{2}

    Now multiply the results of the two brackets.

  9. Multiply the coefficients

    4×9=364 \times 9 = 36

    Four times nine is thirty-six.

  10. Add the x indices

    x6+4=x10x^{6+4} = x^{10}

    Six plus four is ten, so the numerator is 36x to the power 10 y squared.

  11. Rewrite the whole fraction

    36x10y212x5y\frac{36x^{10}y^{2}}{12x^5 y}

    Now divide by twelve x to the power 5 y.

  12. Divide the coefficients

    36÷12=336 \div 12 = 3

    Thirty-six divided by twelve is three.

  13. Divide the x terms

    x10−5=x5x^{10-5} = x^{5}

    Ten minus five is five.

  14. Divide the y terms

    y2−1=y1y^{2-1} = y^{1}

    The denominator y has index 1, so 2 minus 1 gives 1, which is just y.

  15. State the answer

    3x5y3x^{5}y

    The fully simplified expression is three x to the power 5 y.

Answer
3x5y3x^{5}y
Question 2
5 markschallenging
A student claims that x2+x3=x5x^2 + x^3 = x^5. Which statement is correct?
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Worked solution

  1. Read the student's claim

    x2+x3=x5x^2 + x^3 = x^5

    The student has added the indices 2 and 3 to get 5.

  2. Recall when indices are added

    am×an=am+na^m \times a^n = a^{m+n}

    Indices are only added when powers are multiplied, not when terms are added.

  3. Identify the operation here

    x2+x3x^2 + x^3

    This is an addition of two different terms, not a multiplication.

  4. Check whether the terms are alike

    x2 and x3x^2 \text{ and } x^3

    The terms have different indices, so they are not like terms and cannot be combined.

  5. Test the claim with a number

    let x=2\text{let } x = 2

    Substitute a value to test whether the claim could be true.

  6. Work out the left-hand side

    22+23=4+8=122^2 + 2^3 = 4 + 8 = 12

    Two squared plus two cubed is four plus eight, which is twelve.

  7. Work out the claimed answer

    25=322^5 = 32

    Two to the power 5 is thirty-two.

  8. Compare the two values

    12≠3212 \neq 32

    Twelve does not equal thirty-two, so the claim is false.

  9. Explain why

    addition≠multiplication\text{addition} \neq \text{multiplication}

    You may only add indices when multiplying powers of the same base.

  10. Consider the multiplication case

    x2×x3=x5x^2 \times x^3 = x^{5}

    If it had been a product, the answer x to the power 5 would be correct.

  11. State what can be done

    x2+x3=x2(1+x)x^2 + x^3 = x^2(1 + x)

    The sum can only be factorised, not written as a single power.

  12. Rule out the multiply-indices idea

    x2⋅x3≠x6x^2 \cdot x^3 \neq x^{6}

    Multiplying the indices to get 6 is also wrong; indices are added when multiplying.

  13. Rule out the doubling idea

    x2+x3≠2x5x^2 + x^3 \neq 2x^5

    The two terms are not equal, so they cannot be counted as two of the same term.

  14. Confirm the correct statement

    cannot be simplified to a single power\text{cannot be simplified to a single power}

    The expression x squared plus x cubed cannot be written as one power of x.

  15. Choose the answer

    option describing that indices add only when multiplying\text{option describing that indices add only when multiplying}

    The correct statement explains that indices are added only when multiplying.

Answer
x2+x3 cannot be written as a single powerx^2 + x^3\text{ cannot be written as a single power}
Question 3
5 markschallenging
Simplify (x12×x52x)2\left(\frac{x^{\frac{1}{2}} \times x^{\frac{5}{2}}}{x}\right)^{2}.
Show worked solution

Worked solution

  1. Look at the numerator inside first

    x12×x52x^{\frac{1}{2}} \times x^{\frac{5}{2}}

    Simplify the top of the inner fraction before dividing.

  2. Recall the multiplication law

    am×an=am+na^m \times a^n = a^{m+n}

    Add the indices even when they are fractions.

  3. Add the fractional indices

    x12+52x^{\frac{1}{2}+\frac{5}{2}}

    The indices share a denominator, so add the numerators.

  4. Simplify the sum

    12+52=62=3\frac{1}{2}+\frac{5}{2} = \frac{6}{2} = 3

    One half plus five halves is six halves, which is 3.

  5. Write the inner numerator

    x3x^{3}

    The numerator simplifies to x cubed.

  6. Rewrite the inner fraction

    x3x\frac{x^{3}}{x}

    Now divide x cubed by x, which has index 1.

  7. Recall the division law

    am÷an=am−na^m \div a^n = a^{m-n}

    Subtract the indices when dividing.

  8. Subtract the indices

    x3−1x^{3-1}

    Three minus one.

  9. Simplify the inside

    x2x^{2}

    Three minus one is two, so the bracket contains x squared.

  10. Apply the outer power

    (x2)2(x^{2})^{2}

    Now raise x squared to the power 2.

  11. Recall the power-of-a-power law

    (am)n=amn(a^m)^n = a^{mn}

    Multiply the indices.

  12. Multiply the indices

    x2×2x^{2 \times 2}

    Two times two.

  13. Simplify the power

    x4x^{4}

    Two times two is four.

  14. Check with a value

    at x=4: (2×324)2=162=256=44\text{at } x=4:\ \left(\frac{2 \times 32}{4}\right)^2 = 16^2 = 256 = 4^{4}

    Using x=4x = 4 gives 256, which is 4 to the power 4.

  15. State the answer

    x4x^{4}

    The fully simplified expression is x to the power 4.

Answer
x4x^{4}
Question 4
5 markschallenging
Simplify 4x3(2x2)3\frac{4x^3}{(2x^2)^3}. Give your answer with a positive index.
Show worked solution

Worked solution

  1. Deal with the denominator bracket first

    (2x2)3(2x^2)^3

    Apply the power 3 to the bracket in the denominator.

  2. Apply the power to each factor

    23×(x2)32^3 \times (x^2)^3

    The power 3 applies to both the 2 and the x squared.

  3. Cube the coefficient

    23=82^3 = 8

    Two cubed is eight.

  4. Multiply the indices

    x2×3=x6x^{2 \times 3} = x^{6}

    Two times three is six, so the denominator is 8x to the power 6.

  5. Rewrite the fraction

    4x38x6\frac{4x^{3}}{8x^{6}}

    Now divide 4x cubed by 8x to the power 6.

  6. Divide the coefficients

    48=12\frac{4}{8} = \frac{1}{2}

    Four over eight simplifies to one half.

  7. Recall the division law

    am÷an=am−na^m \div a^n = a^{m-n}

    Subtract the indices when dividing.

  8. Subtract the indices

    x3−6x^{3-6}

    Three minus six.

  9. Work out the index

    x−3x^{-3}

    Three minus six is negative three.

  10. Write the current form

    12x−3\frac{1}{2}x^{-3}

    The coefficient is one half with a negative power of x.

  11. Recall the negative-index law

    a−n=1ana^{-n} = \frac{1}{a^{n}}

    A negative index means take the reciprocal.

  12. Rewrite the negative power

    x−3=1x3x^{-3} = \frac{1}{x^{3}}

    x to the power negative 3 becomes 1 over x cubed.

  13. Combine the fractions

    12×1x3\frac{1}{2} \times \frac{1}{x^{3}}

    Multiply the one half by 1 over x cubed.

  14. Simplify the product

    12x3\frac{1}{2x^{3}}

    Multiplying the denominators gives two x cubed underneath.

  15. State the answer

    12x3\frac{1}{2x^{3}}

    The answer with a positive index is 1 over two x cubed.

Answer
12x3\frac{1}{2x^{3}}
Question 5
5 markschallenging
Simplify (x4y6x−2y2)12\left(\frac{x^4 y^{6}}{x^{-2} y^{2}}\right)^{\frac{1}{2}}.
Show worked solution

Worked solution

  1. Simplify inside the bracket first

    x4y6x−2y2\frac{x^4 y^{6}}{x^{-2} y^{2}}

    Deal with the division before applying the square root.

  2. Set up the x subtraction

    x4−(−2)x^{4-(-2)}

    Subtract the denominator x index, which is negative 2.

  3. Handle the double negative

    4−(−2)=4+2=64-(-2) = 4+2 = 6

    Subtracting negative 2 gives 6, so the x term is x to the power 6.

  4. Divide the y terms

    y6−2=y4y^{6-2} = y^{4}

    Six minus two is four.

  5. Write the simplified inside

    x6y4x^{6} y^{4}

    The bracket now contains x to the power 6 y to the power 4.

  6. Interpret the fractional power

    (x6y4)12=x6y4(x^6 y^4)^{\frac{1}{2}} = \sqrt{x^6 y^4}

    A power of one half means the square root.

  7. Apply the power to each factor

    (x6)12×(y4)12(x^6)^{\frac{1}{2}} \times (y^4)^{\frac{1}{2}}

    The power one half applies to both letters.

  8. Recall the power-of-a-power law

    (am)n=amn(a^m)^n = a^{mn}

    Multiply the indices for each power.

  9. Halve the x index

    x6×12x^{6 \times \frac{1}{2}}

    Multiply 6 by one half.

  10. Simplify the x power

    x3x^{3}

    Half of six is three.

  11. Halve the y index

    y4×12y^{4 \times \frac{1}{2}}

    Multiply 4 by one half.

  12. Simplify the y power

    y2y^{2}

    Half of four is two.

  13. Combine the parts

    x3y2x^{3} y^{2}

    Bring the two powers together.

  14. Check by squaring

    (x3y2)2=x6y4(x^3 y^2)^2 = x^6 y^4

    Squaring x cubed y squared returns the inside of the bracket.

  15. State the answer

    x3y2x^{3}y^{2}

    The fully simplified expression is x cubed y squared.

Answer
x3y2x^{3}y^{2}

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