Factorization of Quadratic Expression of the Form ax2 + bx c where a, b, c are constant.
Mathematics SSS 1 Second Term
WEEK 1
Factorization of Quadratic Expression of the Form ax2 + bxc where a, b, c are constant.
Performance Objectives
Students should be able to;
- Factorize quadratic expressions.
- Solve quadratic equations of the form ab = 0, i.e. a = 0 or b = 0.
- Construct quadratic equations with given roots
- Solve problems involving quadratic equations.
Revision of Linear and quadratic expressions
Content
Any expression in which the highest power of the unknown is 1 is called a linear expression.
Examples
a) x + 1
b) 2y + 3
c) p = −1/2
In general, linear expressions are expressions of the form ax + b, where a & b are constants and x is a variable.
A quadratic expression is that whose highest power of the unknown is 2.
Examples
1. x2 + 3x
2. 2x2 − 6x + 10
Factorization of quadratic expressions:
Examples:
1. Factorize the following quadratic expressions
(a. x2 + 4x
(b. 2x2 − 8x
Solution:
(a. x2 + 4x
x is a common factor of the terms x2 and 4x. Hence x2 + 4x can be written as x. x2 + 4x. Isolating common factors, we have, x(x2 + 4)
(b. 2x2 − 8x
The common factor of the terms 2x2 and 6x is 2x
∴ 2x2 − 8x can be written as 2x. x − 2x. 4,
hence we obtain 2x(x − 4)
2. Factorize the following:
(a. x2 + 8x − 20
(b. 6a2 + 15a + 9
(c. 7 − 22x + 3x2
Solution:
(a. x2 + 8x − 20
Find the product of the first and last terms
x2 × (−20) = −20x2
Find two terms such that their product is −20x2 and their sum is +8x
|
Factors of −20x2 |
Sum of factors |
|
|
1. |
−20x and +x |
−19x |
|
2. |
+20x and −x |
+19x |
|
3. |
−10x and +2x |
−8x |
|
4. |
+10x and −2x |
+8x |
|
5. |
−5x and +4x |
−x |
|
6. |
+5x and −4x |
+x |
Of these, only 4 gives the required result. Replace +8x with +10x and −2x in the given expression. Then factorize by grouping the terms.
x2 + 8x − 20
= x2 + 10x − 2x − 20
= x(x + 10) − 2(x + 10)
= (x + 10)(x − 2)
(b. 6a2 + 15a + 9, 3 is common factor, first take out the common factor.
3(2a2 + 5a + 3)
2a2 × 3 = 6a2
|
Factors of +6a2 |
Sum of factors |
|
|
(a. |
+6a and +a |
+2a |
|
(b. |
+3a and +2a |
+5a |
6a2 + 15a + 9
= 3(2a2 + 5a + 3)
= 3(2a2 + 3a + 2a + 3)
= 3[a(2a + 3) + 1(2a + 3)]
= 3(2a + 3)(a + 1)
(c. 7 − 22x + 3x2
Find the product of the first and last terms i.e 7 × (+3x2 ) = +21x2
Find two terms such that their sum is −22x and their product is +21x2. Since the middle term is negative, consider negative factors only. The terms are −21x and −x. Replace −22x with −21x −x in the given expression.
7 − 22x + 3x2
= 7 − 21x − x + 3x2
= 7(1 − 3x) −x(1 − 3x)
= (1 − 3x)(7 − x)
Solution of Quadratic Expression of The Form
ab = 0, a = 0 OR b = 0
If the product of two numbers is 0, then one of the numbers (or possibly both of them) must be zero.
For example, 3 × 0 = 0, 0 × 5 = 0 and 0 × 0 = 0
In general, if a × b = 0, then either a = 0 or b = 0 or both a and b are zero.
Examples
Solve the equation (x − 2)(x + 7) = 0
Solution
If (x − 2)(x + 7) = 0, the either (x − 2) = 0 or (x + 7) = 0
⇒ x = 2 or x = −7
2. Solve the equation a(a + 3) = 0
Solution
If a(a + 3) = 0, then either a = 0 or a + 3 = 0
⇒ a = 0 or a = −3
3. Solve the equations (i) (2m − 5)2 = 0 (ii) d(d − 4 )(d + 6)2 = 0
Solution:
(i) If (2m − 5)2 = 0
Then, (2m − 5)(2m − 5) = 0
(2m − 5) = 0 twice
⇒ m = 5/2 twice
(ii) If d(d − 4 )(d + 6)2 = 0, then any one of the four factors of LHS may be 0
i.e. d = 0, d − 4 = 0, (d + 6)2 = 0
⇒ d = 0, d = 4, or d = −6 twice
Formation of Quadratic Equation with Given Roots
The roots of a quadratic equation are the solutions of that equation. Suppose the roots of a quadratic equation in x are a and b, then we can write; x = a and x = b
Examples;
1. Find the quadratic equation whose roots are −2 and +2
Solution:
Let x = −2 or x = 2, then
x + 2 = 0 or x − 2 = 0
(x + 2)(x − 2) = 0
On careful expansion, we obtain x2 − 4 = 0
Find the quadratic equation whose roots are 2½ and −1
Solution:
If the roots are 2½ and −1
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Let x = 2½ and x = −1 X = 5/2 and x = −1 2x = 5 and x = −12x – 5 = 0 or x+1=0 ⇒ (2x − 5)(x + 1) = 0 2x2 + 2x − 5x – 5 = 0 2x2 − 3x – 5 = 0 |