Factoring Quadratics
Factoring Quadratics
A quadratic expression in maths is an expression whose highest power or degree is `2`. A quadratic expression in its standard form is written as `ax^2 + bx + c`. Here
`a` is the coefficient of the `x^2` (squared) term
`b` is the coefficient of the `x` (linear) term
`c` is the constant term
Factoring quadratics means breaking down a quadratic expression into simpler parts, called linear factors. This helps us understand and solve quadratic equations more easily. By factoring quadratics, we can find the roots of the equation, which are the points where a quadratic graph, also called parabola, crosses the `x`-axis.
It's an important tool in math that simplifies complicated quadratic equations by factorizing them into linear factors. Let's explore how this process works with some examples.
What Is Factoring Quadratics?
Factoring quadratics means breaking down a quadratic equation into its linear factors, like `(x - k)(x - h)`, where `h` and `k` are the roots of the equation. The linear factors and the roots of a quadratic are related by factor theorem. The theorem states that if `x = a` and `x = b` are two roots of the quadratic then `(x - a)` and `(x - b)` should be the two factors of the quadratic. This method is also known as the factorization of quadratic equations. There are various ways to factorize quadratics, such as splitting the middle term, using the quadratic formula, or completing the square.We can employ different strategies to factorize quadratics, depending on the equation's structure.
Factorization Methods of Quadratic Equations
Factorization of quadratic equations can be approached in various ways. Some common methods include:
`1`. Splitting the middle term: This technique involves breaking down the middle term (the linear term) of the quadratic expression into two terms whose product equals the product of the first and last terms.
`2`. Using a formula: Certain quadratic equations follow specific formulas for factorization, making the process more straightforward.
`3`. Using the Quadratic formula: In cases where other methods are impractical, the quadratic formula provides a reliable way to find the roots of a quadratic equation.
`4`. Using algebraic identities: Leveraging identities like the difference of squares or perfect square trinomials can simplify the factorization process.
Factoring Quadratics by Extracting GCF
Factoring quadratics by extracting the greatest common factor (GCF) is an effective method when all the terms in the quadratic expression have a common factor other than `1`. This method simplifies the quadratic to a more manageable form and is often a preliminary step before applying other factoring techniques like factoring trinomials, using the quadratic formula, or completing the square.
Example: Factorize the quadratic expression `3x^2 + 15x`.
Solution:
The two terms of the expression `3x^2` and `15x` have the factors `3` and `x` common in both of them.
Hence the GCF of the expression `3x^2 + 15x` is `3x`.
After extracting GCF from both the terms we get `x + 5x`.
Hence we can write `3x^2 + 15x` in factored form as `3x(x + 5)`.
Factoring Quadratics by Splitting the Middle Term
To factorize a quadratic equation by splitting the middle term, follow these steps:
Step `1`: Begin with the quadratic equation `ax^2 + bx + c = 0`.
Step `2`: Find two numbers whose product is equal to the product of "`a`" and "`c`", and whose sum equals "`b`".
Step `3`: Split the middle term of the quadratic equation using these two numbers.
Step `4`: Take out the common factors from each pair and simplify the expression.
Let's illustrate this with an example:
Example: Factorize the quadratic equation `2x^2 + 5x + 2 = 0` by splitting the middle term.
Solution:
Given equation: `2x^2 + 5x + 2 = 0`
Here, `a = 2, b = 5, c = 2`
Product of `a` and `c = (2)(2) = 4`
Factors of `4: 1, 2, 4`
Let's find two factors whose sum is `5`:
Sum of two factors `= 5 = 1 + 4`
Now, split the middle term using the factors `1` and `4`.
`2x^2 + x + 4x + 2 = 0`
Take out the common terms:
`x(2x + 1) + 2(2x + 1) = 0`
`(x + 2)(2x + 1) = 0`
Thus, `(x + 2)` and `(2x + 1)` are the factors of the given quadratic equation.
Solving these factors, we find `x = -2, -1/2` as roots.
Factoring Quadratics Using Formula
To factorize a quadratic equation using the formula method, follow these steps:
Step `1`: Begin with the quadratic equation `ax^2 + bx + c = 0`.
Step `2`: Find two numbers whose product is equal to "`a`" times "`c`" and whose sum equals "`b`".
Step `3`: Substitute these two numbers into the formula: `(1/a) [ax + (\text{number}\ 1)] [ax + (\text{number}\ 2)] = 0`.
Step `4`: Simplify the equation obtained in the previous step.
Let's understand this method better with an example:
Example: Factorize `2x^2 + 5x + 2 = 0`.
Solution:
Given equation: `2x^2 + 5x + 2 = 0`
Here, `a = 2, b = 5, c = 2`
Product of `a` and `c = (2)(2) = 4`
Factors of `4: 1, 2, 4`
Let's find two factors whose sum is `5`:
Sum of two factors `= 5 = 1 + 4`
Product of these two factors `= (1)(4) = 4`
Now, substitute these two numbers into the formula: `(1/2) [2x + 1] [2x + 4] = 0`.
`(1/2) (2x + 1)(2x + 4) = 0`
`(2x + 1)(2x + 4) = 0`
Now, simplify the equation:
`2(2x + 1)(x + 2) = 0`
`(2x + 1)(x + 2) = 0`
Thus, `(2x + 1)` and `(x + 2)` are the factors of the given quadratic equation.
By solving these factors, we find `x = -1/2, -2` as roots.
Quadratic Formula for Factoring Quadratics
Factoring quadratics can also be done using a quadratic formula that provides the roots of the quadratic equation and, consequently, its factors.
If `ax^2 + bx + c = 0` is a quadratic equation, where "`a`" is the coefficient of `x^2`, "`b`" is the coefficient of `x`, and "`c`" is the constant term, then the quadratic formula to find the value(s) of `x` is:
`x = (-b ± sqrt(b² - 4ac)) / (2a)`
Example: Factorize `x^2 + 6x + 8 = 0`.
Solution:
Here, `a = 1, b = 6,` and `c = 8`.
Substituting these values into the formula, we get:
`x = (-6 ± sqrt(6^2 - 4 * 1 * 8)) / (2 * 1)`
`= (-6 ± sqrt(36 - 32)) / 2`
`= (-6 ± sqrt4) / 2`
`= (-6 ± 2) / 2`
`x = (-6 + 2) / 2` and `x = (-6 - 2) / 2`
This gives us two roots: `-2` and `-4`
Thus, the factors are `(x + 2)` and `(x + 4)`.
Factoring Quadratics Using Special Cases
We can also use special cases like perfect square trinomials and difference of squares to factorize quadratic equations.
Factoring Quadratics Using Perfect Square Trinomials
The expansion for the two perfect square trinomials are:
\( (a + b)^2 = a^2 + 2ab + b^2 \)
\( (a - b)^2 = a^2 - 2ab + b^2 \)
When we have a quadratic equation in the form of ` a^2 + 2ab + b^2 ` or ` a^2 - 2ab + b^2 `, we can factor it directly using these identities.
Example `1`: Factorize `x^2 + 8x + 16 = 0`.
Solution:
We can rewrite ` x^2 + 8x + 16 = 0 ` as ` (x)^2 + 2(4)(x) + (4)^2 = 0 `, which is in the form ` a^2 + 2ab + b^2 `.
As per perfect square trinomial, we know ` a^2 + 2ab + b^2 = (a + b)^2 `.
Hence, we can write ` x^2 + 8x + 16 = 0 ` as ` (x + 4)^2 = 0 `, which simplifies to ` (x + 4)(x + 4) = 0 `. Therefore, the factors are ` (x + 4) ` and ` (x + 4) `.
Example `2`: Factorize `x^2 - 10x + 25 = 0`.
Solution:
We can rewrite ` x^2 - 10x + 25 = 0 ` as ` (x)^2 - 2(5)(x) + (5)^2 = 0 `, which is in the form ` a^2 - 2ab + b^2 `.
As per perfect square trinomial, we know ` a^2 - 2ab + b^2 = (a - b)^2 `.
Hence, we can write ` x^2 - 10x + 25 = 0 ` as ` (x - 5)^2 = 0 `, which simplifies to ` (x - 5)(x - 5) = 0 `.
Thus, the factors are ` (x - 5) ` and ` (x - 5) `.
Factoring Quadratics Using Difference of Squares
As per the difference of square formula ` a^2 - b^2 = (a + b)(a - b) `.
If the given quadratics is written in the form of difference of two square values, we can use this method so factorize such quadratics.
Example: Factorize ` 4x^2 - 9 = 0 `.
Solution:
We can rewrite `4x^2 - 9 = 0 ` as ` (2x)^2 - (3)^2 = 0 `, which is in the form ` a^2 - b^2 `.
As per difference of squares formula, we know, ` a^2 - b^2 = (a + b)(a - b) `.
Hence, it factors as ` (2x + 3)(2x - 3) = 0 `.
Solved Examples
Example `1`: Factorize the quadratic equation ` 4x^2 + 8x + 4 = 0 `.
Solution:
Given equation: ` 4x^2 + 8x + 4 = 0 `
Rewrite as: ` (2x)^2 + 2(2x)(2) + (2)^2 = 0 `
Identifying it as ` a^2 + 2ab + b^2 `, it factors as ` (2x + 2)^2 = 0 `
Which simplifies to ` (2x + 2)(2x + 2) = 0 `
Thus, the factors are ` (2x + 2) ` and ` (2x + 2) `.
Example `2`: Factorize the quadratic equation ` 9x^2 - 25 = 0 `.
Solution:
Given equation: ` 9x^2 - 25 = 0 `
This can be rewritten as ` (3x)^2 - 5^2 = 0 `
Recognizing it as ` a^2 - b^2 `, it factors as ` (3x + 5)(3x - 5) = 0 `
Thus, the factors are ` (3x + 5) ` and ` (3x - 5) `.
Example `3`: Factorize the quadratic equation ` x^2 + 6x = 0 `.
Solution:
The two terms of the expression `x^2` and `6x` have the factor `x` common in both of them.
Hence the GCF of the expression `x^2 + 6x` is `x`.
After extracting GCF from both the terms we get `x + 6`.
Hence we can write `x^2 + 6x` in factored form as `x(x + 6)`.
Thus, the factored form of a quadratic equation ` x^2 + 6x = 0 ` is ` x(x + 6) = 0`.
Example `4`: Factorize the quadratic equation ` 2x^2 - 3x - 5 = 0 `.
Solution:
Given equation: ` 2x^2 - 3x - 5 = 0 `
Using the quadratic formula:
`x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}`
\( a = 2, \space b = -3, \space c = -5 \)
`x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}`
`x = \frac{3 \pm \sqrt{9 + 40}}{4}`
`x = \frac{3 \pm \sqrt{49}}{4}`
`x = \frac{3 \pm 7}{4}`
`x = \frac{10}{4} \text{or } x = \frac{-4}{4}`
Thus, the roots are ` x = \frac{5}{2} ` and ` x = -1 `.
Example `5`: Factorize the quadratic equation ` 6x^2 - 7x + 2 = 0 `.
Solution:
Given equation: ` 6x^2 - 7x + 2 = 0 `
Splitting the middle term, we get:
\( 6x^2 - 3x - 4x + 2 = 0 \)
Factor by grouping:
\( 3x(2x - 1) - 2(2x - 1) = 0 \)
\( (3x - 2)(2x - 1) = 0 \)
The factored quadratic equation is ` (3x - 2)(2x - 1) = 0 `.
Thus, the factors are ` (3x - 2) ` and ` (2x - 1) `.
Practice Problems
Q`1`. Factorize the quadratic equation ` 4x^2 + 12x + 9 = 0 `.
- ` (2x + 3)^2 `
- ` (2x + 3)(2x + 3) `
- ` (2x - 3)^2 `
- ` (2x - 3)(2x - 3) `
Answer: a
Q`2`. Factorize the quadratic equation ` x^2 - 7x + 12 = 0 `.
- ` (x - 4)(x - 3) `
- ` (x + 4)(x - 3) `
- ` (x - 4)(x + 3) `
- ` (x + 4)(x + 3) `
Answer: a
Q`3`. Factoring a quadratic equation using the quadratic formula always results in integer factors. (True/False)
- True
- False
Answer: b
Q`4`. What are the linear factors of the quadratic equation ` 6x^2 - 11x + 4 = 0 `?
- ` (2x - 1)(3x + 4) `
- ` (2x + 4)(x - 3) `
- ` (2x - 1)(3x - 4) `
- ` (3x - 1)(2x - 4) `
Answer: c
Q`5`. Which two factors multiply to produce the quadratic expression ` x^2 - 9 `?
- ` (x + 3)(x + 3) `
- ` (x - 3)(x + 3) `
- ` (x - 3)(x - 3) `
- ` (x - 4)(x - 5) `
Answer: b
Frequently Asked Questions
Q`1`. What is factoring in mathematics?
Answer: Factoring in mathematics refers to the process of breaking down a mathematical expression or number into its constituent factors, which when multiplied together result in the original expression or number. In algebra, factoring involves finding the factors of a polynomial expression, typically to simplify it or to solve equations.
Q`2`. What is factorisation of quadratic equation?
Answer: Factoring quadratics involves rewriting a quadratic expression, typically in the form `ax^2+bx+c` as a product of two binomials or other simpler expressions.
Q`3`. What does it mean to factor by extracting the GCF?
Answer: Factoring by extracting the GCF (Greatest Common Factor) involves identifying the largest common factor in the coefficients of the terms in a quadratic expression and pulling it out as a common factor.
Q`4`. How to factor a quadratic equation?
Answer: Quadratic equations can be factored using various methods, including:
`1`. Factoring by splitting the middle term: Finding factors of `a` times `c` that add up to `b`.
`2`. Factoring by grouping: Grouping terms in a quadratic expression and finding common factors.
`3`. Factoring special forms: Recognizing patterns such as perfect squares or the difference of squares.
`4`. Factoring using the quadratic factoring formula: Using the quadratic formula to find the roots of the equation, helps determine the factors.
`5`. Factoring by completing the square: Transforming the quadratic expression into a perfect square trinomial and factoring it.
Q`5`. What are the applications of factoring in real life?
Answer: Factoring has numerous applications in real life, including:
`1`. Finance: Factoring helps in financial analysis, such as calculating interest rates, loan payments, and investment returns.
`2`. Engineering: Factoring is used in engineering calculations, such as determining dimensions, forces, and stresses in structures.
`3`. Computer science: Factoring is essential in cryptography for secure communication and data encryption.
`4`. Economics: Factoring is used in economic modeling and analysis, such as forecasting trends and analyzing market behavior.
`5`. Science: Factoring plays a role in scientific research and experimentation, such as modeling physical phenomena and analyzing experimental data.