Subsection 1.3.3 Factorisation Of Quadratic Expressions.
Factorisation is the process of breaking down a quadratic expression into a product of simpler binomials.
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.
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.
When the coefficient of \(x^2\) is one, the expression is of the form;
\begin{equation*}
x^2 + bx + c.
\end{equation*}
Cosider the following expressions:
\begin{equation*}
\text{(a).} \, (x + 3)(x + 4).
\end{equation*}
\begin{equation*}
\text{(b).} \, (x -6)(x - 5).
\end{equation*}
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*}
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*}
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.
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 \(b\text{.}\)
In each case;
The sum of the constant terms in the factors is equal to the coefficient of \(x\) in the expression.
The product of the constant terms in the factors is equal to the constant term in the expression.
Example 1.3.10.
Factorise the following expression:
\begin{equation*}
x^2 + 5x + 6.
\end{equation*}
Solution.
In this case, the coefficient of \(x^2\) is one, the coefficient of \(x\) is 5 and the constant term is 6.
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.
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.
In this case, the numbers are 2 and 3.
\begin{equation*}
x^2 + 5x + 6 = x^2 + 2x + 3x + 6.
\end{equation*}
Grouping terms, we get:
\begin{equation*}
x^2 + 5x + 6 = x(x + 2) + 3(x + 2).
\end{equation*}
\begin{equation*}
x^2 + 5x + 6 = (x + 2)(x + 3).
\end{equation*}
Therefore, the factorised form of the expression \(x^2 + 5x + 6\) is \((x + 2)(x + 3)\text{.}\)
Example 1.3.11.
Factorise the following expression:
\begin{equation*}
x^2 - 9x + 20.
\end{equation*}
Solution.
Look for two numbers such that the numbers \(a\) and \(b\) such that \(a + b = -9\) and \(ab = 20\text{.}\)
The numbers are -4 and -5.
Then, the expression \(x^2 - 9x + 20\) can be written as:
\begin{equation*}
x^2 - 4x - 5x + 20.
\end{equation*}
Grouping terms, we get:
\begin{equation*}
x^2 - 9x + 20 = x(x - 4) - 5(x - 4).
\end{equation*}
\begin{equation*}
x^2 - 9x + 20 = (x - 4)(x - 5).
\end{equation*}
The factorised form of the expression \(x^2 - 9x + 20\) is \((x - 4)(x - 5)\text{.}\)
Example 1.3.12.
Factorise the following expression:
\begin{equation*}
x^2 + 6x + 9.
\end{equation*}
Solution.
Look for two numbers such that the numbers \(a\) and \(b\) such that \(a + b = 6\) and \(ab = 9\text{.}\)
The numbers are 3 and 3.
Then, the expression \(x^2 + 6x + 9\) can be written as:
\begin{equation*}
x^2 + 3x + 3x + 9.
\end{equation*}
Grouping terms, we get:
\begin{equation*}
x^2 + 6x + 9 = x(x + 3) + 3(x + 3).
\end{equation*}
\begin{equation*}
x^2 + 6x + 9 = (x + 3)(x + 3).
\end{equation*}
The factorised form of the expression \(x^2 + 6x + 9\) is \((x + 3)(x + 3)\text{.}\)
Exercises Exercises
1.
Factorise the following expressions:
\(\displaystyle x^2 + 4x + 4.\)
\(\displaystyle x^2 + 8x + 15.\)
\(\displaystyle x^2 - 7x + 12.\)
\(\displaystyle x^2 - 6x + 9.\)
\(\displaystyle x^2 + 3x + 2.\)
\(\displaystyle x^2 - 5x + 6.\)
\(\displaystyle x^2 + 2x - 15.\)
\(\displaystyle x^2 - 4x - 5.\)
\(\displaystyle x^2 - 3x - 10.\)
\(\displaystyle x^2 + 7x + 10.\)
\(\displaystyle x^2 - 8x - 20.\)
\(\displaystyle x^2 + 9x + 20.\)
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.
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 \(1\text{.}\)
Consider the following identities:
-
\((3x + 2)(2x + 1)\)
\begin{equation*}
= (3x \times 2x) + (3x \times 1) + (2 \times 2x) + (2 \times 1)
\end{equation*}
\begin{equation*}
= 6x^2 + 3x + 4x + 2
\end{equation*}
\begin{equation*}
= 6x^2 + 7x + 2
\end{equation*}
-
\((3x - 2)(4x - 3)\)
\begin{equation*}
= (3x \times 4x) - (3x \times 3) - (2 \times 4x) + (2 \times 3)
\end{equation*}
\begin{equation*}
= 12x^2 - 9x - 8x + 6
\end{equation*}
\begin{equation*}
= 12x^2 - 17x + 6
\end{equation*}
-
\((2x + 3)(3x - 4)\)
\begin{equation*}
= (2x \times 3x) - (2x \times 4) + (3 \times 3x) - (3 \times 4)
\end{equation*}
\begin{equation*}
= 6x^2 - 8x + 9x - 12
\end{equation*}
\begin{equation*}
= 6x^2 + x - 12
\end{equation*}
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.
Factors are multiplied to get the final expresions on the righthand side (RHS) of the equations.
Investigation 1.3.9.
Given the quadratic expressions on the right-hand side of the equation, how can you be able factor them?
Consider the expression \(6x^2 + 7x + 2\text{.}\)
The problem can factorised as follows:
Look for two numbers such that:
-
Their product is equal to the product of the coefficient of \(x^2\) and the constant term, i.e. \(6 \times 2 = 12\text{.}\)
In this case \(6\) is the coefficient of \(x^2\) and \(2\) is constant term.
-
Their sum is \(7\text{,}\) where \(7\) is the coefficient of \(x\text{.}\)
The terms are \(3\) and \(4\)
-
Rewrite the term \(7x\) as the sum of the two numbers found in step (i)(b).
Thus,
\begin{equation*}
6x^2 + 7x + 2 = 6x^2 + 3x + 4x + 2
\end{equation*}
\begin{equation*}
= 3x(2x + 1) + 2(2x + 1)
\end{equation*}
\begin{equation*}
= (3x + 2)(2x + 1)
\end{equation*}
Example 1.3.13.
Factorise the expression \(12x^2 - 17x + 6\text{.}\)
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*}
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*}
Splt the middle term using these two numbers:
\begin{equation*}
12x^2 - 9x - 8x + 6
\end{equation*}
Factor by grouping:
\begin{equation*}
3x(4x - 3) - 2(4x - 3)
\end{equation*}
Group the factors:
\begin{equation*}
(3x - 2)(4x - 3)
\end{equation*}
Example 1.3.14.
Factorise the expression \(3x^2 - 5x - 2\text{.}\)
Solution.
The expression can be factorised as follows:
Look for two numbers such that:
-
Their product is equal to the product of the coefficient of \(x^2\) and the constant term, i.e. \(3 \times -2 = -6\text{.}\)
In this case \(3\) is the coefficient of \(x^2\) and \(-2\) is constant term.
-
Their sum is \(-5\text{,}\) where \(-5\) is the coefficient of \(x\text{.}\)
The terms are \(-6\) and \(1\)
-
Rewrite the term \(-5x\) as the sum of the two numbers found in step 1.
Thus,
\begin{equation*}
3x^2 - 5x - 2 = 3x^2 - 6x + x - 2
\end{equation*}
\begin{equation*}
= 3x(x - 2) + 1(x - 2)
\end{equation*}
\begin{equation*}
= (3x + 1)(x - 2)
\end{equation*}
Exercises Exercises
1.
Factorise the following expressions:
\(\displaystyle 3x^2 + 4x - 7\)
\(\displaystyle 5x^2 - 9x + 4\)
\(\displaystyle 11x^2 - 15x + 10\)
\(\displaystyle 17x^2 - 25x + 16\)
\(\displaystyle 2x^2 + 7x + 3\)
\(\displaystyle 13x^2 - 17x + 12\)
\(\displaystyle 6x^2 + 5x - 4\)
\(\displaystyle 4x^2 + 19x - 13\)
\(\displaystyle 7x^2 - 13x + 6\)
\(\displaystyle 8x^2 + 3x - 5\)
\(\displaystyle 9x^2 - 14x + 8\)
\(\displaystyle 10x^2 + 11x - 6\)
\(\displaystyle 15x^2 - 21x + 14\)
\(\displaystyle 4x^2 - 11x + 6\)
\(\displaystyle 12x^2 + 13x - 11\)
\(\displaystyle 16x^2 + 23x - 15\)
Subsection 1.3.5 Solutions Of Quadratic Equations By Factorisations.
Activity 1.3.11.
Work in Groups.
Find the roots of the equation using factorization.
\(x² - 5x + 6 = 0\text{.}\)
\(x² + 7x + 10 = 0\text{.}\)
After solving the equations, discuss the following:
What steps did you follow in factorizing the quadratic equation?
How do the roots relate to the factors of the quadratic equation?
Did any group use a different method to factorize the equation? Compare approaches.
Finally, each group will share their solved quadratic equations and the methods they used infront of the classroom.
In our previous sections we have discussed the form of a quadratic expression, identities and quadratic equations; how to form quadratic equations and differentiate between an expression and equation
In this section we are going to learn about solving the factored equation, we are going to continue from the point where we form factored equation
Numbers that satisfy an equation (its solutions) are called the roots of the equation.
Once you have factored the quadratic into the form;
\begin{align*}
(x+p)(x+q) \amp = 0
\end{align*}
Set each factor equal to zero and solve for \(x\text{:}\)
\begin{align*}
(x + p) \amp = 0
\end{align*}
\begin{align*}
\text{or} \amp
\end{align*}
\begin{align*}
(x + q) \amp = 0
\end{align*}
Solving these will give the two solutions for \(x\text{.}\)
Example 1.3.17.
Solve
\begin{align}
x^2 + 5x + 6 \amp = 0 \tag{1.3.1}
\end{align}
Solution.
Look for two numbers that if multiplied by \(6\) and when added up gives \(5\)
These numbers are \(2\) and \(3.\) So we can write the equation as:
\begin{align*}
x^2 + 2x + 3x + 6 \amp = 0
\end{align*}
Finding a common factor we have:
\begin{align*}
x(x + 2) + 3(x + 2) \amp = 0
\end{align*}
Grouping we have th factors of the equation as:
\begin{align*}
(x +2) (x + 3) \amp = 0
\end{align*}
Setting each factor equal to zero (0).
\begin{equation*}
x + 2 = 0
\end{equation*}
\begin{equation*}
\text{or}
\end{equation*}
\begin{equation*}
x + 3 = 0
\end{equation*}
Solving for \(x.\)
\begin{equation*}
x = -2
\end{equation*}
\begin{equation*}
\text{or}
\end{equation*}
\begin{equation*}
x = -3
\end{equation*}
Thus, the solutions to the quadratic equation \(x^+5x+6=0\) are;
\begin{equation*}
x = (-2) \, \text {and} \, x = (-3).
\end{equation*}
Example 1.3.18.
Solve,
\begin{align}
6x^2 + 13x + 6 \amp = 0 \tag{1.3.2}
\end{align}
Solution.
We need 2 numbers to form factors, which when;
These numbers are \(9\) and \(4.\) So rewrite the middle part.
\begin{align*}
6x^2 + 9x + 4x + 6 \amp = 0
\end{align*}
Grouping the factors:
\begin{align*}
(6x^2 + 9x) + (4x + 6) \amp = 0
\end{align*}
Factor each group:
\begin{align*}
3x(2x + 3) + 2(2x + 3) \amp = 0
\end{align*}
\(\,\)
\begin{align*}
(2x+3)(3x+2) \amp = 0
\end{align*}
Set each factor eaqual to zero:
\begin{equation*}
2x + 3 = 0
\end{equation*}
\begin{equation*}
\text{or}
\end{equation*}
\begin{equation*}
3x + 2 = 0
\end{equation*}
Solving for \(x.\)
\begin{equation*}
x = -\frac{3}{2}
\end{equation*}
\begin{equation*}
\text{or}
\end{equation*}
\begin{equation*}
x = -\frac{2}{3}
\end{equation*}
Thus, the solutions to \(6x^2 +13x+6=0\) are:
\begin{equation*}
x = -\frac{3}{2} \, \text{and} \, x = -\frac{2}{3}
\end{equation*}
Exercises Exercises
1.
Solve the quadratic equation by factorisation.
\(\displaystyle x^2 + 7x + 10 = 0\)
\(\displaystyle x^2 - 5x + 6 = 0\)
\(\displaystyle x^2 + 3x - 4 = 0\)
\(\displaystyle 2x^2 + 9x + 7 = 0\)
\(\displaystyle x^2 - 6x + 8 = 0\)
2.
A car’s speed is represented by a quadratic equation:
\begin{align*}
4x^2 - 16x + 15 \amp = 0 \text{.}
\end{align*}
Find the possible values of x representing time.
The sum of a number and its square is \(42\text{.}\) Form a quadratic equation and solve it to find the number.
Solve the quadratic equation:
\begin{equation*}
3x^2 - 14x + 8 = 0
\end{equation*}
A garden’s area is 56 square meters, and its length is 4 meters more than its width. Form and solve a quadratic equation to find the dimensions of the garden.
Subsection 1.3.6 Application Of Quadratic Equations To Real Life Situations.
Quadratic equations are a type of mathematical equation that can be used to describe many different real-life situations. Whether you’re throwing a ball in the air, trying to maximize the area of a garden, or calculating profits for a business, quadratic equations can help solve problems that involve relationships with squared terms.
In this section, we will learn how to recognize real-life situations that can be modeled by quadratic equations, how to set them up, and how to solve them step by step.
A quadratic equation is an equation that can be written in the form:
\begin{equation*}
ax^2 + bx + c = 0
\end{equation*}
where,
\(a\text{,}\) \(b\) and \(c\) are real numbers or constants,
\(x\) is the unknown variable
and \(a \neq 0\) beacuse if \(a\) is \(0,\) it would npt be a quadratic
Quadratic equations show up in many situations, especially when something is changing in a way that involves squares (like height, area, or profit).
Lets take a look at some of the situations Quadratic Equations is involved in real life situations.
\(1. \, Projectile \, Motion (Throwing \, a \, Ball).\)
One of the most common real-life situations for quadratic equations is projectile motion, such as when you throw a ball in the air. The height of the ball over time can be described by a quadratic equation.
Example 1.3.19.
A rock is dropped from a height of \(50\) meters. Its height above the ground at time \(t\) is given by
\begin{equation*}
h(t) = 5t^2 + 50.
\end{equation*}
Use factorization to determine how long it will take for the rock to reach the ground.
Solution.
The height of the rock above the ground is given by:
\begin{equation*}
h(t) = -5t^2 + 50
\end{equation*}
To find when the rock reaches the ground, set \(h(t) = 0\text{:}\)
\begin{equation*}
-5t^2 + 50 = 0
\end{equation*}
Simplify the equation:
\begin{equation*}
-5t^2 = -50
\end{equation*}
\begin{equation*}
t^2 = 10
\end{equation*}
Solve for \(t\text{:}\)
\begin{equation*}
t = \pm \sqrt{10}
\end{equation*}
Since time cannot be negative, \(t = \sqrt{10}\text{.}\)
Therefore, the rock will take approximately \(\sqrt{10} \approx 3.16\) seconds to reach the ground.
\(2. \, Maximizing \, Area \, (Optimization \, Problem).\)
Quadratic equations are also used in optimization problems. These are problems where you want to maximize or minimize something, like the area of a garden.
Example 1.3.20.
The length of a rectangle is \(3\) meters more than its width. If the area of the rectangle is \(54\) square meters, find the dimensions of the rectangle by factoring the quadratic equation.
Solution.
Let the width of the rectangle be \(x\text{.}\) Then the length is \(x + 3\text{.}\)
The area of the rectangle is given by:
\begin{equation*}
x(x + 3) = 54
\end{equation*}
Expand and rearrange into standard quadratic form:
\begin{equation*}
x^2 + 3x - 54 = 0
\end{equation*}
Factorize the quadratic equation:
\begin{equation*}
(x + 9)(x - 6) = 0
\end{equation*}
Solve for \(x\text{:}\)
\begin{equation*}
x = -9 \, \text{or} \, x = 6
\end{equation*}
Since width cannot be negative, \(x = 6\text{.}\)
The dimensions of the rectangle are:
\begin{equation*}
\text{Width} = 6 \, \text{meters}, \, \text{Length} = 6 + 3 = 9 \, \text{meters}.
\end{equation*}
Since the width cannot be negative, the width is \(6\) meters, and lenght is \(6 + 3 = 9\) meters.
\(3. Business \, Applications (Profit \, Function)\)
Businesses often use quadratic equations to model their profit. For example, the profit a company makes from selling a product can be modeled by a quadratic equation.
Example 1.3.21.
A company’s profit \(P(x)\) from selling \(x\) units of a product is given by the equation:
\begin{equation*}
P(x) = -x^2 + 15x - 50
\end{equation*}
Find how many units the company needs to sell to have no profit (i.e., when the profit is zero).
Solution.
To find when the company has no profit, set \(P(x) = 0\text{:}\)
\begin{equation*}
-x^2 + 20x - 50 = 0
\end{equation*}
Multiply through by \(-1\) to simplify:
\begin{equation*}
x^2 - 20x + 50 = 0
\end{equation*}
Factorize the quadratic equation:
\begin{equation*}
(x - 10)(x - 5) = 0
\end{equation*}
Solve for \(x\text{:}\)
\begin{equation*}
x = 10 or x = 5
\end{equation*}
Therefore, the company will have no profit when it sells either \(5\) units or \(10\) units.
Exercises Exercises
1.
A stone is thrown into the air from a height of \(4\) meters with an initial velocity of \(8\) meters per second. The height of the stone at time \(t\) is given by:
\begin{equation*}
h(t) = -5t^2 + 8t + 4
\end{equation*}
Find when the stone reaches the ground.
2.
A farmer has \(200\) meters of fencing. He wants to build a rectangular garden. The length is \(50\) meters longer than the width. What should the dimensions be to maximize the area?
3.
A school’s profit function is given by:
\begin{equation*}
P(x) = -x^2 + 30x - 100
\end{equation*}
Find the number of units the company must sell to achieve zero profit.
4.
A water fountain shoots water into the air, and its height at any time \(t\) (in seconds) is given by the equation:
\begin{equation*}
h(t) = -4.9t^2 + 15t + 2
\end{equation*}
Find the time it will take for the water to return to the ground (i.e., when \(h(t)=0\)).
5.
Solve the following quadratic equation:
\begin{equation*}
x^2 - 7x + 12 = 0
\end{equation*}
Find the value of \(x.\)
6.
A company finds that the revenue \(R(x)\) it generates from selling \(x\) units of a product is given by the quadratic equation:
\begin{equation*}
R(x) = 2x^2 + 40x
\end{equation*}
Find the number of units \(x\) that the company needs to sell to maximize its revenue.
7.
The path of a car is represented by the quadratic equation
\begin{equation*}
y = 2x^2 - 8x,
\end{equation*}
where \(x\) represents time in seconds and \(y\) represents the car’s position. Find the time when the car reaches its starting position by factorizing the equation.
