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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 1.3.3 Factorisation Of Quadratic Expressions.

Factorisation is the process of breaking down a quadratic expression into a product of simpler binomials.
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Subsubsection 1.3.3.1 Factorisation When The Coefficient Of \(x^2\) Is One

The expression
\begin{equation*} ax^2 + bx + c, \end{equation*}
where \(a, b, c\) are constants and \(a \neq 0\text{,}\) is called a quadratic expression.
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In such expressions \(a\) is called the coefficient of \(x^2\text{,}\) \(b\) is called the coefficient of \(x\) and \(c\) is called the constant term.
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When the coefficient of \(x^2\) is one, the expression is of the form;
\begin{equation*} x^2 + bx + c. \end{equation*}
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Cosider the following expressions:
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\begin{equation*} \text{(a).} \, (x + 3)(x + 4). \end{equation*}
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\begin{equation*} \text{(b).} \, (x -6)(x - 5). \end{equation*}
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Expanding the expressions we get:
\begin{align*} (x + 3)(x + 4) \amp = x(x + 3) + 4(x + 3) \end{align*}
\begin{align*} (x + 3)(x + 4) \amp = x^2 + 3x + 4x + 12 \end{align*}
Collecting like terms to get:
\begin{equation*} x^2 + 7x + 12. \end{equation*}
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Expanding the expressions we get:
\begin{align*} (x - 6)(x - 5) \amp = x(x - 5) - 6(x - 5) \end{align*}
\begin{align*} (x - 6)(x - 5) \amp = x^2 - 5x - 6x + 30 \end{align*}
Simplify the middle terms:
\begin{equation*} x^2 - 11x + 30. \end{equation*}
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From the above examples, we can see that the expressions \(x^2 + 7x + 12\) and \(x^2 - 11x + 30\) are formed from the factors of expressions \((x + 3)(x + 4)\) and \((x - 6)(x - 5)\) respectively.
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The factorised form of the expression \(x^2 + bx + c\) is \((x + m)(x + n)\) where \(m\) and \(n\) are the factors of \(c\) whose sum is \(b\text{.}\)
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In each case;
  1. The sum of the constant terms in the factors is equal to the coefficient of \(x\) in the expression.
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  2. The product of the constant terms in the factors is equal to the constant term in the expression.
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Example 1.3.10.

Factorise the following expression:
\begin{equation*} x^2 + 5x + 6. \end{equation*}
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Solution.
In this case, the coefficient of \(x^2\) is one, the coefficient of \(x\) is 5 and the constant term is 6.
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So the factors will be of the form \((x + m)(x + n)\) where \(m\) and \(n\) are the factors of 6 whose sum is 5.
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In the expression \(x^2 + 5x + 6\text{,}\) look for two numbers such the numbers \(a\) and \(b\) such that
\begin{equation*} a + b = 5 \end{equation*}
is coefficient of \(x\) and \(ab = 6\) is the constant term.
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In this case, the numbers are 2 and 3.
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\begin{equation*} x^2 + 5x + 6 = x^2 + 2x + 3x + 6. \end{equation*}
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Grouping terms, we get:
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\begin{equation*} x^2 + 5x + 6 = x(x + 2) + 3(x + 2). \end{equation*}
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\begin{equation*} x^2 + 5x + 6 = (x + 2)(x + 3). \end{equation*}
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Therefore, the factorised form of the expression \(x^2 + 5x + 6\) is \((x + 2)(x + 3)\text{.}\)
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Example 1.3.11.

Factorise the following expression:
\begin{equation*} x^2 - 9x + 20. \end{equation*}
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Solution.
Look for two numbers such that the numbers \(a\) and \(b\) such that \(a + b = -9\) and \(ab = 20\text{.}\)
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The numbers are -4 and -5.
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Then, the expression \(x^2 - 9x + 20\) can be written as:
\begin{equation*} x^2 - 4x - 5x + 20. \end{equation*}
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Grouping terms, we get:
\begin{equation*} x^2 - 9x + 20 = x(x - 4) - 5(x - 4). \end{equation*}
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\begin{equation*} x^2 - 9x + 20 = (x - 4)(x - 5). \end{equation*}
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The factorised form of the expression \(x^2 - 9x + 20\) is \((x - 4)(x - 5)\text{.}\)
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Example 1.3.12.

Factorise the following expression:
\begin{equation*} x^2 + 6x + 9. \end{equation*}
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Solution.
Look for two numbers such that the numbers \(a\) and \(b\) such that \(a + b = 6\) and \(ab = 9\text{.}\)
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The numbers are 3 and 3.
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Then, the expression \(x^2 + 6x + 9\) can be written as:
\begin{equation*} x^2 + 3x + 3x + 9. \end{equation*}
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Grouping terms, we get:
\begin{equation*} x^2 + 6x + 9 = x(x + 3) + 3(x + 3). \end{equation*}
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\begin{equation*} x^2 + 6x + 9 = (x + 3)(x + 3). \end{equation*}
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The factorised form of the expression \(x^2 + 6x + 9\) is \((x + 3)(x + 3)\text{.}\)
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Exercises Exercises

1.
Factorise the following expressions:
  1. \(\displaystyle x^2 + 4x + 4.\)
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  2. \(\displaystyle x^2 + 8x + 15.\)
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  3. \(\displaystyle x^2 - 7x + 12.\)
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  4. \(\displaystyle x^2 - 6x + 9.\)
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  5. \(\displaystyle x^2 + 3x + 2.\)
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  6. \(\displaystyle x^2 - 5x + 6.\)
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  7. \(\displaystyle x^2 + 2x - 15.\)
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  8. \(\displaystyle x^2 - 4x - 5.\)
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  9. \(\displaystyle x^2 - 3x - 10.\)
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  10. \(\displaystyle x^2 + 7x + 10.\)
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  11. \(\displaystyle x^2 - 8x - 20.\)
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  12. \(\displaystyle x^2 + 9x + 20.\)
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Subsubsection 1.3.3.2 Factorisation When The Coefficient Of \(x^2\) Is Not One.

The general form of a quadratic expression is \(ax^2 + bx + c\text{,}\) where \(a\text{,}\) \(b\) and \(c\) are constants.
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In this section we are going to discuss how to factorise a quadratic expression when the \(a\) in \(ax^2 + bx + c\) is not equal to \(1\text{.}\)
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Consider the following identities:
  1. \((3x + 2)(2x + 1)\)
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    \begin{equation*} = (3x \times 2x) + (3x \times 1) + (2 \times 2x) + (2 \times 1) \end{equation*}
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    \begin{equation*} = 6x^2 + 3x + 4x + 2 \end{equation*}
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    \begin{equation*} = 6x^2 + 7x + 2 \end{equation*}
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  2. \((3x - 2)(4x - 3)\)
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    \begin{equation*} = (3x \times 4x) - (3x \times 3) - (2 \times 4x) + (2 \times 3) \end{equation*}
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    \begin{equation*} = 12x^2 - 9x - 8x + 6 \end{equation*}
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    \begin{equation*} = 12x^2 - 17x + 6 \end{equation*}
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  3. \((2x + 3)(3x - 4)\)
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    \begin{equation*} = (2x \times 3x) - (2x \times 4) + (3 \times 3x) - (3 \times 4) \end{equation*}
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    \begin{equation*} = 6x^2 - 8x + 9x - 12 \end{equation*}
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    \begin{equation*} = 6x^2 + x - 12 \end{equation*}
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From the above examples, we can see that the factorisation of a quadratic expression when the coefficient of \(x^2\) is not one is similar to the factorisation of a quadratic expression when the coefficient of \(x^2\) is one.
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Factors are multiplied to get the final expresions on the righthand side (RHS) of the equations.
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Investigation 1.3.9.

Given the quadratic expressions on the right-hand side of the equation, how can you be able factor them?
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Consider the expression \(6x^2 + 7x + 2\text{.}\)
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The problem can factorised as follows:
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  1. Look for two numbers such that:
    1. Their product is equal to the product of the coefficient of \(x^2\) and the constant term, i.e. \(6 \times 2 = 12\text{.}\)
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      In this case \(6\) is the coefficient of \(x^2\) and \(2\) is constant term.
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    2. Their sum is \(7\text{,}\) where \(7\) is the coefficient of \(x\text{.}\)
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      The terms are \(3\) and \(4\)
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  2. Rewrite the term \(7x\) as the sum of the two numbers found in step (i)(b).
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    Thus,
    \begin{equation*} 6x^2 + 7x + 2 = 6x^2 + 3x + 4x + 2 \end{equation*}
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    \begin{equation*} = 3x(2x + 1) + 2(2x + 1) \end{equation*}
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    \begin{equation*} = (3x + 2)(2x + 1) \end{equation*}
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Example 1.3.13.

Factorise the expression \(12x^2 - 17x + 6\text{.}\)
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Solution.
Multiply the coefficient of \(x^2\) (which is 12) by the constant term (which is 6):
\begin{equation*} 12 \times 6 = 72 \end{equation*}
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Find two numbers that multiply to give \(72\) and add up to \(-17\text{.}\) The two numbers are \(-8\) and \(-9\text{.}\)
\begin{equation*} -9 \times -8 = 72 \end{equation*}
and
\begin{equation*} -9 + -8 = -17 \end{equation*}
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Splt the middle term using these two numbers:
\begin{equation*} 12x^2 - 9x - 8x + 6 \end{equation*}
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Factor by grouping:
\begin{equation*} 3x(4x - 3) - 2(4x - 3) \end{equation*}
Group the factors:
\begin{equation*} (3x - 2)(4x - 3) \end{equation*}
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Example 1.3.14.

Factorise the expression \(3x^2 - 5x - 2\text{.}\)
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Solution.
The expression can be factorised as follows:
  1. Look for two numbers such that:
    1. Their product is equal to the product of the coefficient of \(x^2\) and the constant term, i.e. \(3 \times -2 = -6\text{.}\)
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      In this case \(3\) is the coefficient of \(x^2\) and \(-2\) is constant term.
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    2. Their sum is \(-5\text{,}\) where \(-5\) is the coefficient of \(x\text{.}\)
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      The terms are \(-6\) and \(1\)
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  2. Rewrite the term \(-5x\) as the sum of the two numbers found in step 1.
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    Thus,
    \begin{equation*} 3x^2 - 5x - 2 = 3x^2 - 6x + x - 2 \end{equation*}
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    \begin{equation*} = 3x(x - 2) + 1(x - 2) \end{equation*}
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    \begin{equation*} = (3x + 1)(x - 2) \end{equation*}
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Exercises Exercises

1.
Factorise the following expressions:
    1. \(\displaystyle 3x^2 + 4x - 7\)
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    2. \(\displaystyle 5x^2 - 9x + 4\)
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    3. \(\displaystyle 11x^2 - 15x + 10\)
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    4. \(\displaystyle 17x^2 - 25x + 16\)
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    1. \(\displaystyle 2x^2 + 7x + 3\)
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    2. \(\displaystyle 13x^2 - 17x + 12\)
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    3. \(\displaystyle 6x^2 + 5x - 4\)
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    4. \(\displaystyle 4x^2 + 19x - 13\)
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    1. \(\displaystyle 7x^2 - 13x + 6\)
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    2. \(\displaystyle 8x^2 + 3x - 5\)
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    3. \(\displaystyle 9x^2 - 14x + 8\)
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    4. \(\displaystyle 10x^2 + 11x - 6\)
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    1. \(\displaystyle 15x^2 - 21x + 14\)
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    2. \(\displaystyle 4x^2 - 11x + 6\)
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    3. \(\displaystyle 12x^2 + 13x - 11\)
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    4. \(\displaystyle 16x^2 + 23x - 15\)
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