Algebraic Expressions

Free

Grade 11

1. Completing the Square

Completing the square rewrites a quadratic expression ax²+bx+c into the form a(x+p)²+q, which reveals the turning point of the related parabola directly. The method involves: factoring out a from the x² and x terms, then adding and subtracting (half the coefficient of x)² inside the bracket to create a perfect square.

Colour key:factored-out coefficientcompleted square termconstant adjustment

Method steps

Factor out a — If a ≠ 1, factor out a from the x² and x terms only. Leave the constant term outside the bracket.

Half and square — Take half the coefficient of x inside the bracket, square it, then add and subtract that value inside the bracket.

Simplify — Write the first three terms inside the bracket as a perfect square, then collect the constant terms outside.

Key identity

x²+bx = (x + b/2)² − (b/2)² — adding and subtracting (b/2)² keeps the expression equal while creating the perfect square.

Worked Examples

1
Take half of 6 (the x coefficient): 6 ÷ 2 = 3. Square it: 3² = 9.
2
Add and subtract 9 inside the expression: x²+6x+9−9+5.
3
The first three terms form a perfect square: (x+3)². Collect the constants: −9+5 = −4.
4
Result: x²+6x+5 = (x+3)²−4 ✓

Answer(x+3)²−4

1
Factor out 2 from the first two terms: 2(x²−4x)+3.
2
Inside the bracket: half of −4 is −2; squared is 4. Add and subtract: 2(x²−4x+4−4)+3.
3
Write as a perfect square and expand the subtracted term outside: 2(x−2)²−2×4+3.
4
Simplify the constants: −8+3 = −5.
5
Result: 2x²−8x+3 = 2(x−2)²−5 ✓

Answer2(x−2)²−5

1
Half of −10 is −5. Squared: (−5)² = 25.
2
Add and subtract: x²−10x+25−25+30 = (x−5)²+5.
3
His answer is (x−5)²+5, which matches. Sipho is correct ✓

AnswerSipho is correct

2. Finding Maximum or Minimum Values Using Completing the Square

Once a quadratic expression is written in completed square form a(x+p)²+q, the minimum value (if a>0) or maximum value (if a<0) occurs at x = −p, giving the minimum/maximum value of q itself, since (x+p)² is always ≥ 0.

Colour key:completed square formturning point x-valuemax/min value

Key rules

a > 0 → Minimum

The parabola opens upward. The expression has a minimum value of q at x = −p.

The squared term can only add to q, never subtract from it.

a < 0 → Maximum

The parabola opens downward. The expression has a maximum value of q at x = −p.

The squared term is multiplied by a negative, so it can only reduce q.

Why this works

(x+p)² ≥ 0 always. When a>0: a(x+p)² ≥ 0, so a(x+p)²+q ≥ q — the smallest possible value is q, reached when x = −p. When a<0: a(x+p)² ≤ 0, so the largest possible value is q.

Worked Examples

1
The completed square form is (x+3)²−4.
2
Since a = 1 > 0, the parabola opens upward — this is a minimum.
3
The minimum occurs at x = −3 (where the squared term equals zero).
4
At x = −3: (−3+3)² − 4 = 0 − 4 = −4.
5
Result: Minimum value is −4, occurring at x = −3 ✓

AnswerMinimum value is −4, occurring when x = −3

1Write the area expression: A = x(12−2x) = 12x−2x².
2Rewrite in standard form: A = −2x²+12x.
3
Factor out −2 from both terms: A = −2(x²−6x).
4
Complete the square inside the bracket: half of −6 is −3; squared is 9. So A = −2(x²−6x+9−9) = −2(x−3)²+18.
5
Since a = −2 < 0, this is a maximum.
6
The maximum occurs at x = 3 and the maximum value is 18 m² ✓

AnswerMaximum area = 18 m², occurring when x = 3 m

Diagram

Parabola sketches side by side: one opening upward (a>0) with minimum at turning point labelled q in green and x-value in orange; one opening downward (a<0) with maximum labelled similarly

3. Simplifying Algebraic Fractions with Binomial Denominators

To simplify, add or subtract algebraic fractions with binomial (two-term) denominators, we factorise where possible, find the LCM of the denominators, then combine into a single fraction and simplify.

Colour key:factorised formsLCMsimplified result

Strategy

Factorise — Factorise every numerator and denominator fully. Look for common factors, difference of squares, and trinomials.

Find the LCM — The LCM of the denominators is the product of all distinct factors. Use it as the common denominator.

Combine and simplify — Convert each fraction to the common denominator, add or subtract the numerators, then cancel any common factors.

Important restriction

Always state the values of x for which the expression is undefined (where any denominator equals zero). These are called excluded values. For example, if a denominator is (x+3), state x ≠ −3.

Worked Examples

1
Factorise the numerator using the difference of squares identity: x²−9 = (x−3)(x+3).
2
Rewrite the fraction: (x−3)(x+3)/(x+3).
3
Cancel the common factor (x+3): x−3, where x ≠ −3.
4
Result: (x²−9)/(x+3) = x−3 ✓

Answerx−3

1Identify the denominators: (x+1) and (x−2). Both are already fully factorised.
2
Find the LCM: (x+1)(x−2).
3
Convert each fraction: 2/(x+1) = 2(x−2)/[(x+1)(x−2)] and 3/(x−2) = 3(x+1)/[(x+1)(x−2)].
4
Add the numerators: [2(x−2)+3(x+1)] / [(x+1)(x−2)].
5
Expand the numerator: 2x−4+3x+3 = 5x−1.
6
Result: (5x−1)/[(x+1)(x−2)], where x ≠ −1 and x ≠ 2 ✓

Answer(5x−1)/[(x+1)(x−2)]

1
Factorise the denominator of the first fraction using the difference of squares: x²−16 = (x−4)(x+4).
2
Rewrite: (x+4)/[(x−4)(x+4)] − 1/(x−4).
3
Cancel (x+4) from the first fraction: 1/(x−4) − 1/(x−4).
4
Both fractions are now identical. Subtract: 0, where x ≠ ±4.
5
Result: (x+4)/(x²−16) − 1/(x−4) = 0 ✓

Answer0

Next: Equations and Inequalities

Back to Grade 11

Algebraic Expressions

Free

Grade 11

1. Completing the Square

Colour key:factored-out coefficientcompleted square termconstant adjustment

Method steps

Factor out a — If a ≠ 1, factor out a from the x² and x terms only. Leave the constant term outside the bracket.

Half and square — Take half the coefficient of x inside the bracket, square it, then add and subtract that value inside the bracket.

Simplify — Write the first three terms inside the bracket as a perfect square, then collect the constant terms outside.

Key identity

x²+bx = (x + b/2)² − (b/2)² — adding and subtracting (b/2)² keeps the expression equal while creating the perfect square.

Worked Examples

1
Take half of 6 (the x coefficient): 6 ÷ 2 = 3. Square it: 3² = 9.
2
Add and subtract 9 inside the expression: x²+6x+9−9+5.
3
The first three terms form a perfect square: (x+3)². Collect the constants: −9+5 = −4.
4
Result: x²+6x+5 = (x+3)²−4 ✓

Answer(x+3)²−4

1
Factor out 2 from the first two terms: 2(x²−4x)+3.
2
Inside the bracket: half of −4 is −2; squared is 4. Add and subtract: 2(x²−4x+4−4)+3.
3
Write as a perfect square and expand the subtracted term outside: 2(x−2)²−2×4+3.
4
Simplify the constants: −8+3 = −5.
5
Result: 2x²−8x+3 = 2(x−2)²−5 ✓

Answer2(x−2)²−5

1
Half of −10 is −5. Squared: (−5)² = 25.
2
Add and subtract: x²−10x+25−25+30 = (x−5)²+5.
3
His answer is (x−5)²+5, which matches. Sipho is correct ✓

AnswerSipho is correct

2. Finding Maximum or Minimum Values Using Completing the Square

Colour key:completed square formturning point x-valuemax/min value

Key rules

a > 0 → Minimum

The parabola opens upward. The expression has a minimum value of q at x = −p.

The squared term can only add to q, never subtract from it.

a < 0 → Maximum

The parabola opens downward. The expression has a maximum value of q at x = −p.

The squared term is multiplied by a negative, so it can only reduce q.

Why this works

(x+p)² ≥ 0 always. When a>0: a(x+p)² ≥ 0, so a(x+p)²+q ≥ q — the smallest possible value is q, reached when x = −p. When a<0: a(x+p)² ≤ 0, so the largest possible value is q.

Worked Examples

1
The completed square form is (x+3)²−4.
2
Since a = 1 > 0, the parabola opens upward — this is a minimum.
3
The minimum occurs at x = −3 (where the squared term equals zero).
4
At x = −3: (−3+3)² − 4 = 0 − 4 = −4.
5
Result: Minimum value is −4, occurring at x = −3 ✓

AnswerMinimum value is −4, occurring when x = −3

1Write the area expression: A = x(12−2x) = 12x−2x².
2Rewrite in standard form: A = −2x²+12x.
3
Factor out −2 from both terms: A = −2(x²−6x).
4
Complete the square inside the bracket: half of −6 is −3; squared is 9. So A = −2(x²−6x+9−9) = −2(x−3)²+18.
5
Since a = −2 < 0, this is a maximum.
6
The maximum occurs at x = 3 and the maximum value is 18 m² ✓

AnswerMaximum area = 18 m², occurring when x = 3 m

Diagram

Parabola sketches side by side: one opening upward (a>0) with minimum at turning point labelled q in green and x-value in orange; one opening downward (a<0) with maximum labelled similarly

3. Simplifying Algebraic Fractions with Binomial Denominators

Colour key:factorised formsLCMsimplified result

Strategy

Factorise — Factorise every numerator and denominator fully. Look for common factors, difference of squares, and trinomials.

Find the LCM — The LCM of the denominators is the product of all distinct factors. Use it as the common denominator.

Combine and simplify — Convert each fraction to the common denominator, add or subtract the numerators, then cancel any common factors.

Important restriction

Always state the values of x for which the expression is undefined (where any denominator equals zero). These are called excluded values. For example, if a denominator is (x+3), state x ≠ −3.

Worked Examples

1
Factorise the numerator using the difference of squares identity: x²−9 = (x−3)(x+3).
2
Rewrite the fraction: (x−3)(x+3)/(x+3).
3
Cancel the common factor (x+3): x−3, where x ≠ −3.
4
Result: (x²−9)/(x+3) = x−3 ✓

Answerx−3

1Identify the denominators: (x+1) and (x−2). Both are already fully factorised.
2
Find the LCM: (x+1)(x−2).
3
Convert each fraction: 2/(x+1) = 2(x−2)/[(x+1)(x−2)] and 3/(x−2) = 3(x+1)/[(x+1)(x−2)].
4
Add the numerators: [2(x−2)+3(x+1)] / [(x+1)(x−2)].
5
Expand the numerator: 2x−4+3x+3 = 5x−1.
6
Result: (5x−1)/[(x+1)(x−2)], where x ≠ −1 and x ≠ 2 ✓

Answer(5x−1)/[(x+1)(x−2)]

1
Factorise the denominator of the first fraction using the difference of squares: x²−16 = (x−4)(x+4).
2
Rewrite: (x+4)/[(x−4)(x+4)] − 1/(x−4).
3
Cancel (x+4) from the first fraction: 1/(x−4) − 1/(x−4).
4
Both fractions are now identical. Subtract: 0, where x ≠ ±4.
5
Result: (x+4)/(x²−16) − 1/(x−4) = 0 ✓

Answer0

Next: Equations and Inequalities

Back to Grade 11