Solve a quadratic equation by factorising
Key Notes :
Understanding Factoring
- Factoring is the process of breaking down a polynomial expression into simpler expressions (factors) that can be multiplied together to obtain the original expression.
- In the context of quadratic equations, factoring involves finding two binomial expressions that multiply together to give the original equation.
Steps to Solve by Factoring
- Express the Quadratic Equation in Standard Form: Ensure the equation is in the form ax² + bx + c = 0.
- Factor the Quadratic Expression: Find two binomial expressions that multiply together to give the original quadratic expression. This often involves finding two numbers that add up to b and multiply to c.
- Set Each Factor Equal to Zero: Once the equation is factored, set each factor equal to zero.
- Solve for x: Solve each resulting equation to find the values of x that make the factors zero.
Example
Solve: x² – 5x + 6 = 0
- Step 1: The equation is already in standard form.
- Step 2: Factor the equation: (x – 2)(x – 3) = 0
- Step 3: Set each factor equal to zero:
- x – 2 = 0
- x – 3 = 0
- Step 4: Solve each equation:
- x = 2
- x = 3
Key Points
- Factoring is a powerful technique for solving quadratic equations.
- The goal is to find two binomial expressions that multiply to give the original equation.
- Once factored, set each factor equal to zero and solve for x.
- Not all quadratic equations can be factored easily. In such cases, other methods like completing the square or the quadratic formula can be used.
Learn with an example
Solve for w.
w2+19w–20=0
Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.
w=________
Step 1: Factor.
The c term is –20, so you need to find a pair of factors with a product of –20. The b term is 19, so you need to find a pair of factors with a sum of 19. Since the product is negative (–20), you need one factor to be negative and one factor to be positive.
Make a list of the possible factor pairs with a product of –20, and then find the one with a sum of 19.
| Factor Pairs of c = -20 | Sum of factor pairs |
| 1 . -20 = -20 | 1 + -20 = – 19 |
| -1 . 20 = -20 | -1 + 20 = 19 |
| 2 . 10 = -20 | 2 + – 10 = -8 |
| -2 .10 = -20 | -2 + 10 = 8 |
| 4 . -5 = -20 | 4 + -5 = -1 |
| – 4 .5 = -20 | -4 + 5 = 1 |
The factors 20 and –1 have a sum of 19. Use those numbers to factorise w2+19w–20.
w2+19w–20=0
(w+20)(w–1)=0
Step 2: Use the zero product property to solve.
According to the zero product property, if (w+20)(w–1)=0, then w+20 must be 0 or w–1 must be 0. Write the two equations and solve for w.
w+20=0 or w–1=0
w=-20 w=1
The solutions are w=–20 and w=1.
Solve for q.
q2–25q=0
Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.
q=________
Step 1: Factor.
q2–25q=0
q(q–25)=0 Factor out the highest common factor, q
Step 2: Use the zero product property to solve.
According to the zero product property, if q(q–25)=0, then q must be 0 or q–25 must be 0. Write the two equations and solve for q.
q=0 or q–25=0
q=25
The solutions are q=0 and q=25.
Solve for d.
d2+33d=0
Write each solution as an integer, proper fraction, or improper fraction in simplest form. If there are multiple solutions, separate them with commas.
d=________
Step 1: Factor.
d2+33d=0
d(d+33)=0 Factor out the highest common factor, d
Step 2: Use the zero product property to solve.
According to the zero product property, if d(d+33)=0, then d must be 0 or d+33 must be 0. Write the two equations and solve for d.
d=0 or d+33=0
d=–33
The solutions are d=0 and d=–33.
let’s practice!
