Algebraic Expressions

Free

Grade 10

1. Multiplying a Binomial by a Trinomial

To multiply a binomial (two terms) by a trinomial (three terms), we distribute each term of the binomial across every term of the trinomial, then combine like terms. There are always six partial products before simplifying.

Colour key:first binomial termsecond binomial termcombined result

How distribution works

(a + b)(c + d + e) = ac + ad + ae + bc + bd + be

Each term of the binomial multiplies every term of the trinomial — six products in total.

Steps

Distribute the first term — Multiply the first term of the binomial by every term of the trinomial.

Distribute the second term — Multiply the second term of the binomial by every term of the trinomial.

Combine like terms — Collect terms with the same power of x and add their coefficients.

Write your answer in standard form

Arrange the final answer with the highest power first (x³, then x², then x, then the constant). This makes like terms easy to spot and answers easy to check.

Worked Examples

1
Identify the binomial terms: x and +2.
2
Distribute x: x × x² = x³, x × 3x = +3x², x × (−5) = −5x.
3
Distribute +2: +2 × x² = +2x², +2 × 3x = +6x, +2 × (−5) = −10.
4
Write all six products together: x³ + 3x² − 5x + 2x² + 6x − 10.
5
Combine like terms: x³ | (3 + 2)x² = 5x² | (−5 + 6)x = x | −10.
6
Answer: x³ + 5x² + x − 10

Answerx³ + 5x² + x − 10

1
Identify the binomial terms: 2x and −1.
2
Distribute 2x: 2x × x² = 2x³, 2x × (−4x) = −8x², 2x × 3 = +6x.
3
Distribute −1: −1 × x² = −x², −1 × (−4x) = +4x, −1 × 3 = −3.
4
Write all six products together: 2x³ − 8x² + 6x + −x² + 4x − 3.
5
Combine like terms: 2x³ | (−8 − 1)x² = −9x² | (6 + 4)x = 10x | −3.
6
Answer: 2x³ − 9x² + 10x − 3

Answer2x³ − 9x² + 10x − 3

Diagram

Distribution diagram for (x + 2)(x² + 3x − 5) showing six arrows from each binomial term (x and +2) to each trinomial term (x², 3x, −5), colour coded blue for x and orange for +2

2. Factorising Trinomials

Factorising reverses expansion. For x² + bx + c, find two numbers that multiply to give c and add to give b. For ax² + bx + c where a ≠ 1, use the a × c method: find two numbers that multiply to give ac and add to give b, then split the middle term and factor by grouping.

Colour key:a × c valuesplit termsfactored brackets

Method 1 — x² + bx + c (a = 1)

Find two numbers that multiply to c and add to b.

Write the factored form as (x + p)(x + q) where p and q are your two numbers.

Method 2 — ax² + bx + c (a ≠ 1): the a × c method

Calculate a × c. Find two numbers that multiply to give ac and add to give b.

Split the middle term — replace bx with two separate terms using your two numbers.

Group and factor — factor each pair of terms, then factor out the common binomial.

Always check by expanding

After factorising, expand your answer to verify it matches the original expression. This takes seconds and guarantees your answer is correct.

Worked Examples

1Identify a = 1, b = 2, c = −15. Since a = 1, use Method 1.
2
Find two numbers that multiply to −15 and add to 2.
3
Test factor pairs of 15: 1 × 15, 3 × 5. Try 5 and −3: 5 × (−3) = −15 ✓ and 5 + (−3) = 2 ✓.
4
Write the factored form: (x + 5)(x − 3).
5Check: (x + 5)(x − 3) = x² − 3x + 5x − 15 = x² + 2x − 15 ✓

Answer(x + 5)(x − 3)

1Identify a = 3, b = 11, c = 6. Since a ≠ 1, use the a × c method.
2
Calculate a × c = 3 × 6 = 18.
3
Find two numbers that multiply to 18 and add to 11: 9 and 2 — 9 × 2 = 18 ✓, 9 + 2 = 11 ✓.
4
Split the middle term: 3x² + 9x + 2x + 6.
5
Group: (3x² + 9x) + (2x + 6) = 3x(x + 3) + 2(x + 3).
6
Factor out the common binomial: (3x + 2)(x + 3).
7Check: (3x + 2)(x + 3) = 3x² + 9x + 2x + 6 = 3x² + 11x + 6 ✓

Answer(3x + 2)(x + 3)

1Identify a = 2, b = −7, c = −15. Use the a × c method.
2
Calculate a × c = 2 × (−15) = −30.
3
Find two numbers that multiply to −30 and add to −7: −10 and 3 — (−10) × 3 = −30 ✓, (−10) + 3 = −7 ✓.
4
Split the middle term: 2x² + 3x − 10x − 15.
5
Group: (2x² + 3x) + (−10x − 15) = x(2x + 3) − 5(2x + 3).
6
Factor out the common binomial: (2x + 3)(x − 5).
7Check: (2x + 3)(x − 5) = 2x² − 10x + 3x − 15 = 2x² − 7x − 15 ✓

Answer(2x + 3)(x − 5)

Diagram

X-box diagram for 3x² + 11x + 6 showing a×c = 18 at top, b = 11 at bottom, and the two numbers 9 and 2 at the sides, leading to the split and factored form (3x + 2)(x + 3)

3. Factorising by Grouping in Pairs

When an expression has four or more terms with no single common factor across all terms, we group terms in pairs that share a common factor, factor each pair, then factor out the common binomial factor.

Colour key:first pairsecond paircommon binomial

Steps

Group the first pair — place the first two terms together in brackets.

Group the second pair — place the last two terms in brackets. Watch signs carefully.

Factor each pair, then factor out the common binomial from both groups.

Grouping order matters

Both groups must produce the same binomial factor after factoring. If they do not, try rearranging the terms into a different grouping before concluding the expression cannot be factored this way.

Worked Examples

1
Group the terms in pairs: (ax + ay) + (bx + by).
2
Factor the first pair: a(x + y).
3
Factor the second pair: b(x + y).
4
Both pairs contain the common binomial (x + y).
5
Factor out the common binomial: (x + y)(a + b).
6Check: (x + y)(a + b) = ax + bx + ay + by ✓

Answer(x + y)(a + b)

1
Group the terms in pairs: (x³ + 3x²) + (−4x − 12).
2
Factor the first pair: x²(x + 3).
3
Factor the second pair: −4(x + 3). (Factor out −4 so that (x + 3) appears in both groups.)
4
Both pairs contain the common binomial (x + 3).
5
Factor out the common binomial: (x + 3)(x² − 4).
6
Recognise that x² − 4 is a difference of squares: x² − 4 = (x − 2)(x + 2).
7
Answer: (x + 3)(x − 2)(x + 2)

Answer(x + 3)(x − 2)(x + 2)

Diagram

Flowchart for x³ + 3x² − 4x − 12 showing the first pair red, second pair purple, their factored forms, converging on the common binomial factor, and the final fully factored answer in green

4. Simplifying Algebraic Fractions with Monomial Denominators

To simplify, add, or subtract algebraic fractions with monomial (single-term) denominators, find the LCM of the denominators, convert each fraction, then combine and simplify by factorising the numerator where possible.

Colour key:LCMconverted fractionssimplified answer

Steps for adding or subtracting algebraic fractions

Find the LCM of all the denominators.

Convert each fraction — multiply numerator and denominator by what is needed to reach the LCM denominator.

Combine the numerators, then simplify by factorising the numerator where possible.

State restrictions on the variable

The denominator of a fraction can never equal zero. Always state restrictions such as x ≠ 0 when the variable appears in the denominator. This is a required part of a complete answer in Grade 10.

Worked Examples

1
Factor the numerator: 6x² + 9x = 3x(2x + 3).
2
Write the fraction: 3x(2x + 3) / 3x.
3
Cancel the common factor 3x (allowed since x ≠ 0): 2x + 3.
4
Answer: 2x + 3 (x ≠ 0)

Answer2x + 3 (x ≠ 0)

1
Find the LCM of x and 2x: LCM = 2x.
2
Convert the first fraction: 3/x = 6/(2x) (multiply numerator and denominator by 2).
3
The second fraction 5/(2x) already has the LCM as its denominator.
4
Combine: 6/(2x) + 5/(2x) = 11/(2x).
5
Answer: 11/(2x) (x ≠ 0)

Answer11/(2x) (x ≠ 0)

1
Find the LCM of x² and x: LCM = x².
2
The first fraction 4/x² already has the LCM as its denominator.
3
Convert the second fraction: 1/x = x/x² (multiply numerator and denominator by x).
4
Combine: 4/x² − x/x² = (4 − x)/x².
5The numerator 4 − x cannot be factored further.
6
Answer: (4 − x)/x² (x ≠ 0)

Answer(4 − x)/x² (x ≠ 0)

Next: Exponents

Back to Grade 10

Algebraic Expressions

Free

Grade 10

1. Multiplying a Binomial by a Trinomial

Colour key:first binomial termsecond binomial termcombined result

How distribution works

(a + b)(c + d + e) = ac + ad + ae + bc + bd + be

Each term of the binomial multiplies every term of the trinomial — six products in total.

Steps

Distribute the first term — Multiply the first term of the binomial by every term of the trinomial.

Distribute the second term — Multiply the second term of the binomial by every term of the trinomial.

Combine like terms — Collect terms with the same power of x and add their coefficients.

Write your answer in standard form

Arrange the final answer with the highest power first (x³, then x², then x, then the constant). This makes like terms easy to spot and answers easy to check.

Worked Examples

1
Identify the binomial terms: x and +2.
2
Distribute x: x × x² = x³, x × 3x = +3x², x × (−5) = −5x.
3
Distribute +2: +2 × x² = +2x², +2 × 3x = +6x, +2 × (−5) = −10.
4
Write all six products together: x³ + 3x² − 5x + 2x² + 6x − 10.
5
Combine like terms: x³ | (3 + 2)x² = 5x² | (−5 + 6)x = x | −10.
6
Answer: x³ + 5x² + x − 10

Answerx³ + 5x² + x − 10

1
Identify the binomial terms: 2x and −1.
2
Distribute 2x: 2x × x² = 2x³, 2x × (−4x) = −8x², 2x × 3 = +6x.
3
Distribute −1: −1 × x² = −x², −1 × (−4x) = +4x, −1 × 3 = −3.
4
Write all six products together: 2x³ − 8x² + 6x + −x² + 4x − 3.
5
Combine like terms: 2x³ | (−8 − 1)x² = −9x² | (6 + 4)x = 10x | −3.
6
Answer: 2x³ − 9x² + 10x − 3

Answer2x³ − 9x² + 10x − 3

Diagram

Distribution diagram for (x + 2)(x² + 3x − 5) showing six arrows from each binomial term (x and +2) to each trinomial term (x², 3x, −5), colour coded blue for x and orange for +2

2. Factorising Trinomials

Colour key:a × c valuesplit termsfactored brackets

Method 1 — x² + bx + c (a = 1)

Find two numbers that multiply to c and add to b.

Write the factored form as (x + p)(x + q) where p and q are your two numbers.

Method 2 — ax² + bx + c (a ≠ 1): the a × c method

Calculate a × c. Find two numbers that multiply to give ac and add to give b.

Split the middle term — replace bx with two separate terms using your two numbers.

Group and factor — factor each pair of terms, then factor out the common binomial.

Always check by expanding

After factorising, expand your answer to verify it matches the original expression. This takes seconds and guarantees your answer is correct.

Worked Examples

1Identify a = 1, b = 2, c = −15. Since a = 1, use Method 1.
2
Find two numbers that multiply to −15 and add to 2.
3
Test factor pairs of 15: 1 × 15, 3 × 5. Try 5 and −3: 5 × (−3) = −15 ✓ and 5 + (−3) = 2 ✓.
4
Write the factored form: (x + 5)(x − 3).
5Check: (x + 5)(x − 3) = x² − 3x + 5x − 15 = x² + 2x − 15 ✓

Answer(x + 5)(x − 3)

1Identify a = 3, b = 11, c = 6. Since a ≠ 1, use the a × c method.
2
Calculate a × c = 3 × 6 = 18.
3
Find two numbers that multiply to 18 and add to 11: 9 and 2 — 9 × 2 = 18 ✓, 9 + 2 = 11 ✓.
4
Split the middle term: 3x² + 9x + 2x + 6.
5
Group: (3x² + 9x) + (2x + 6) = 3x(x + 3) + 2(x + 3).
6
Factor out the common binomial: (3x + 2)(x + 3).
7Check: (3x + 2)(x + 3) = 3x² + 9x + 2x + 6 = 3x² + 11x + 6 ✓

Answer(3x + 2)(x + 3)

1Identify a = 2, b = −7, c = −15. Use the a × c method.
2
Calculate a × c = 2 × (−15) = −30.
3
Find two numbers that multiply to −30 and add to −7: −10 and 3 — (−10) × 3 = −30 ✓, (−10) + 3 = −7 ✓.
4
Split the middle term: 2x² + 3x − 10x − 15.
5
Group: (2x² + 3x) + (−10x − 15) = x(2x + 3) − 5(2x + 3).
6
Factor out the common binomial: (2x + 3)(x − 5).
7Check: (2x + 3)(x − 5) = 2x² − 10x + 3x − 15 = 2x² − 7x − 15 ✓

Answer(2x + 3)(x − 5)

Diagram

X-box diagram for 3x² + 11x + 6 showing a×c = 18 at top, b = 11 at bottom, and the two numbers 9 and 2 at the sides, leading to the split and factored form (3x + 2)(x + 3)

3. Factorising by Grouping in Pairs

Colour key:first pairsecond paircommon binomial

Steps

Group the first pair — place the first two terms together in brackets.

Group the second pair — place the last two terms in brackets. Watch signs carefully.

Factor each pair, then factor out the common binomial from both groups.

Grouping order matters

Worked Examples

1
Group the terms in pairs: (ax + ay) + (bx + by).
2
Factor the first pair: a(x + y).
3
Factor the second pair: b(x + y).
4
Both pairs contain the common binomial (x + y).
5
Factor out the common binomial: (x + y)(a + b).
6Check: (x + y)(a + b) = ax + bx + ay + by ✓

Answer(x + y)(a + b)

1
Group the terms in pairs: (x³ + 3x²) + (−4x − 12).
2
Factor the first pair: x²(x + 3).
3
Factor the second pair: −4(x + 3). (Factor out −4 so that (x + 3) appears in both groups.)
4
Both pairs contain the common binomial (x + 3).
5
Factor out the common binomial: (x + 3)(x² − 4).
6
Recognise that x² − 4 is a difference of squares: x² − 4 = (x − 2)(x + 2).
7
Answer: (x + 3)(x − 2)(x + 2)

Answer(x + 3)(x − 2)(x + 2)

Diagram

Flowchart for x³ + 3x² − 4x − 12 showing the first pair red, second pair purple, their factored forms, converging on the common binomial factor, and the final fully factored answer in green

4. Simplifying Algebraic Fractions with Monomial Denominators

Colour key:LCMconverted fractionssimplified answer

Steps for adding or subtracting algebraic fractions

Find the LCM of all the denominators.

Convert each fraction — multiply numerator and denominator by what is needed to reach the LCM denominator.

Combine the numerators, then simplify by factorising the numerator where possible.

State restrictions on the variable

Worked Examples

1
Factor the numerator: 6x² + 9x = 3x(2x + 3).
2
Write the fraction: 3x(2x + 3) / 3x.
3
Cancel the common factor 3x (allowed since x ≠ 0): 2x + 3.
4
Answer: 2x + 3 (x ≠ 0)

Answer2x + 3 (x ≠ 0)

1
Find the LCM of x and 2x: LCM = 2x.
2
Convert the first fraction: 3/x = 6/(2x) (multiply numerator and denominator by 2).
3
The second fraction 5/(2x) already has the LCM as its denominator.
4
Combine: 6/(2x) + 5/(2x) = 11/(2x).
5
Answer: 11/(2x) (x ≠ 0)

Answer11/(2x) (x ≠ 0)

1
Find the LCM of x² and x: LCM = x².
2
The first fraction 4/x² already has the LCM as its denominator.
3
Convert the second fraction: 1/x = x/x² (multiply numerator and denominator by x).
4
Combine: 4/x² − x/x² = (4 − x)/x².
5The numerator 4 − x cannot be factored further.
6
Answer: (4 − x)/x² (x ≠ 0)

Answer(4 − x)/x² (x ≠ 0)

Next: Exponents

Back to Grade 10