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Algebra and Graphs - Algebraic manipulation and factorisation

Grade 11A Level

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Factorisation by taking out common factors: Identifying the highest common factor (HCF) of all terms in an expression. For example, in the expression 12x2+8x12x^2 + 8x, the HCF is 4x4x.

Area model showing 4x being factored out of 12x squared and 8x to leave 3x + 2
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Factorising quadratics of the form x2+bx+cx^2 + bx + c: Find two numbers that multiply to give cc and add to give bb. These numbers form the factors (x+p)(x+q)(x + p)(x + q).

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Difference of two squares: Recognizing expressions in the form a2−b2a^2 - b^2 which always factorise to (a+b)(a−b)(a + b)(a - b).

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Simplifying algebraic fractions: Factorise both the numerator and the denominator completely, then cancel out common factors that appear in both.

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Changing the subject of a formula: Using inverse operations to isolate a specific variable. If the variable appears twice, collect those terms on one side and factorise.

📐Formulae

Difference of Two Squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

Perfect Square (Positive): (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Perfect Square (Negative): (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

Quadratic Form: ax2+bx+cax^2 + bx + c factorises to (px+q)(rx+s)(px + q)(rx + s) where pr=apr=a and qs=cqs=c.

Algebraic Fraction Addition: ab±cd=ad±bcbd\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}

💡Examples

Problem 1:

Factorise completely: 6x2−15xy6x^2 - 15xy.

Solution:

3x(2x−5y)3x(2x - 5y)

Explanation:

Identify the Highest Common Factor (HCF) of the numerical coefficients (3) and the variables (xx). Divide each term by 3x3x and place the result inside brackets.

Problem 2:

Simplify the algebraic fraction: x2−92x2+5x−3\frac{x^2 - 9}{2x^2 + 5x - 3}.

Solution:

x−32x−1\frac{x - 3}{2x - 1}

Explanation:

Factorise the numerator using the Difference of Two Squares: (x−3)(x+3)(x-3)(x+3). Factorise the denominator: (2x−1)(x+3)(2x-1)(x+3). Cancel the common factor (x+3)(x+3) from both the top and bottom.

Problem 3:

Make xx the subject of the formula: y=2x+1x−3y = \frac{2x + 1}{x - 3}.

Solution:

x=3y+1y−2x = \frac{3y + 1}{y - 2}

Explanation:

  1. Multiply both sides by (x−3)(x-3) to get y(x−3)=2x+1y(x-3) = 2x+1. 2. Expand: yx−3y=2x+1yx - 3y = 2x + 1. 3. Move all xx terms to one side: yx−2x=3y+1yx - 2x = 3y + 1. 4. Factorise xx: x(y−2)=3y+1x(y - 2) = 3y + 1. 5. Divide by (y−2)(y-2) to isolate xx.

Problem 4:

Factorise by grouping: ax−ay+bx−byax - ay + bx - by.

Solution:

(a+b)(x−y)(a + b)(x - y)

Explanation:

Group the first two terms a(x−y)a(x-y) and the last two terms b(x−y)b(x-y). Since (x−y)(x-y) is common to both groups, factorise it out to get (a+b)(x−y)(a+b)(x-y).

Problem 5:

Calculate the area of the shaded region in terms of xx and factorise the resulting expression completely: a large square has side length 3x+43x + 4 and a smaller square of side length x−2x - 2 is removed from its center.

A large square with side 3x+4 containing a smaller central square with side x-2

Solution:

Area=(3x+4)2−(x−2)2Area = (3x + 4)^2 - (x - 2)^2 Using a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b): Area=[(3x+4)−(x−2)][(3x+4)+(x−2)]Area = [(3x + 4) - (x - 2)][(3x + 4) + (x - 2)] Area=[3x+4−x+2][3x+4+x−2]Area = [3x + 4 - x + 2][3x + 4 + x - 2] Area=(2x+6)(4x+2)Area = (2x + 6)(4x + 2) Factorise out common constants: Area=2(x+3)⋅2(2x+1)Area = 2(x + 3) \cdot 2(2x + 1) Area=4(x+3)(2x+1)Area = 4(x + 3)(2x + 1)

Explanation:

The problem uses the difference of two squares. We identify the side lengths as aa and bb, apply the identity, and then factor out constants from each binomial to factorise 'completely'.

Problem 6:

The area of a rectangle is represented by x2+7x+10x^2 + 7x + 10. If the width is x+2x + 2, find an expression for the length. Use a graph of y=x2+7x+10y = x^2 + 7x + 10 to identify the x-intercepts.

Graph of y = x squared + 7x + 10 showing intercepts at -2 and -5

Solution:

Length=AreaWidth=x2+7x+10x+2Length = \frac{Area}{Width} = \frac{x^2 + 7x + 10}{x + 2} Factorising the numerator: x2+7x+10=(x+2)(x+5)x^2 + 7x + 10 = (x + 2)(x + 5) Length=(x+2)(x+5)x+2=x+5Length = \frac{(x + 2)(x + 5)}{x + 2} = x + 5 The x-intercepts occur where y=0y = 0: (x+2)(x+5)=0  ⟹  x=−2,x=−5(x + 2)(x + 5) = 0 \implies x = -2, x = -5

Explanation:

To find the length, we divide the area by the width. This requires factorising the quadratic expression. The factors (x+2)(x+2) and (x+5)(x+5) relate directly to the roots/x-intercepts of the function.