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YEAR 8 · ALGEBRAIC EXPRESSIONS · 13 MIN

Multiplying/dividing algebraic terms

Multiply and divide simple algebraic terms by handling numerical and variable factors separately.

In this lesson

Before you start
  • Signed multiplication
  • Exponent notation
01 / LEARN

Understand the idea.

Multiply factors

An algebraic term is a product of its coefficient and variable factors. In (3⁢a)⁢(4⁢b), multiply 3×4 and a×b to obtain 12⁢a⁢b. A repeated factor gives a power, so (3⁢x)⁢(4⁢x)=12x2.

Multiplication and addition do different jobs. The sum 3⁢x+4⁢x is 7⁢x, while their product is 12x2. Keep a negative coefficient in brackets when multiplying, and determine the sign as well as the numerical coefficient.

Divide by common factors

To divide terms, write a fraction and identify factors shared by its numerator and denominator. In 15⁢x⁢y5⁢x, the coefficient quotient is 3 and the shared x factor cancels, leaving 3⁢y. Cancellation means division by a nonzero factor.

Keep restrictions from the original denominator even if its variables disappear. Here x≠0 remains required. Only whole factors cancel: the x in x+3 is a term of a sum, so it cannot simply cancel against a denominator x.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A repeated variable

Simplify (3⁢x)⁢(−4⁢x).

Worked solution

  1. 1

    Separate coefficients and variable factors.

    (3⁢x)⁢(−4⁢x)=(3⁢(−4))(x×x)

    Reordering factors preserves a product and makes the two calculations visible.

  2. 2

    Multiply the coefficients.

    3⁢(−4)=−12

    A positive coefficient times a negative coefficient gives a negative product.

  3. 3

    Combine the repeated variable factors.

    x×x=x2(3⁢x)⁢(−4⁢x)=−12x2

    Two factors of the same variable are represented by a square.

Answer

−12x2

EXAMPLE 02

A quotient of terms

Simplify 18a2b6⁢a⁢b for a≠0,b≠0.

Worked solution

  1. 1

    Separate the common factors.

    18a2b6⁢a⁢b=186×a2a×bb

    The denominator is a product and is nonzero for the stated a and b values.

  2. 2

    Simplify each quotient.

    186=3a2a=abb=1

    One a factor and the b factor cancel by division by nonzero quantities.

  3. 3

    Multiply the remaining factors.

    18a2b6⁢a⁢b=3⁢a,a≠0, b≠0

    The simplified expression keeps the restrictions of the original quotient.

Answer

3⁢a

03 / YOUR TURN

Try it yourself.

Check your understanding

Simplify (2⁢a)⁢(5⁢a).

Show answer & reasoning

10a2

Multiply coefficients and variable factors separately: 2×5=10 and a×a=a2. Hence (2⁢a)⁢(5⁢a)=10a2.

Keep practising

6 worksheet questions · 13 marks

Read the questions online
  1. Simplify (4⁢x)⁢(3⁢y).
    2 marks
  2. Simplify (−2⁢a)⁢(6⁢a).
    2 marks
  3. Simplify 21⁢x⁢y7⁢x for x≠0.
    2 marks
  4. Simplify 24m26⁢m for m≠0.
    2 marks
  5. Simplify (3⁢p)⁢(−2⁢q)⁢(2).
    2 marks
  6. Why does 8⁢t2⁢t=4 require t≠0?
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Simplify (4⁢x)⁢(3⁢y).
    2 marks

    Multiply coefficients and retain both variable factors.

    12⁢x⁢y
  2. Simplify (−2⁢a)⁢(6⁢a).
    2 marks

    −2×6=−12 and a×a=a2.

    −12a2
  3. Simplify 21⁢x⁢y7⁢x for x≠0.
    2 marks

    Divide 21 by 7 and cancel x.

    3⁢y
  4. Simplify 24m26⁢m for m≠0.
    2 marks

    Divide coefficients to get 4.

    Cancel one common factor m.

    4⁢m
  5. Simplify (3⁢p)⁢(−2⁢q)⁢(2).
    2 marks

    Numerical product: 3⁢(−2)⁢(2)=−12.

    Keep p and q.

    −12⁢p⁢q
  6. Why does 8⁢t2⁢t=4 require t≠0?
    3 marks

    Cancellation divides by t.

    Division by zero is not allowed.

    At t=0, the original denominator is zero and the quotient is undefined.
Checking saved progress…
Curriculum reference & lesson scope

Australian Curriculum v9 Mathematics with SA curriculum context

Combining like terms; substitution with signed/rational values; multiplying/dividing algebraic terms; expanding a single bracket; factorising common factors; equivalent linear expressions; rearranging simple formulas.

Official curriculum source ↗

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