Unit 10.5 Other Factoring Methods
Factoring by Substitution
Factoring using substitution is a technique used to simplify the factoring process, especially when dealing with complex polynomials. By substituting a part of the expression with a new variable, the problem can be transformed into a simpler form, which can then be factored more easily. After factoring, the substitution is reversed to return to the original variables
Factor by Substitution
Factor the expression (x + 1)2 – 5 (x + 1) + 6
x + 1 is an expression that is common. We can substitute for a variable that is easier to manage
let u = x + 1
u2 – 5u + 6
Now we can factor it like a normal polynomial
(u – 2) (u – 3)
We need to substitute the original value back into u
((x+1)-2) ((x+1)-3)
(x – 1)(x – 2)
Factoring using Completing the Square
Factoring by completing the square is a method used to rewrite a quadratic expression in the form of a perfect square trinomial. This technique is particularly useful for solving quadratic equations.
- Start with the Standard Form:
- Ensure the quadratic expression is in the form ax2 + bx + c = 0
- Isolate the Quadratic and Linear Terms:
- If necessary, divide all terms by a so that the coefficient on x2 is equal to 1.
- Move the constant term to the right side of the equation
- Rewrite as a Perfect Square Trinomial:
- Rewrite the equation so that the left side is factorable. Add that value to both sides. Use the equation (b/2)2
- Simplify:
- Combine and simplify any remaining constants outside the perfect square trinomial.
Completing the Square
Factor 5x2 – 4x -2 = 0
Divide all terms by 5 to make the leading coefficient = 1
x2 – [latex]\frac{4}{5}[/latex]x – [latex]\frac{2}{5}[/latex] = 0
Move the constant term to the right side
x2 – [latex]\frac{4}{5}[/latex]x = [latex]\frac{2}{5}[/latex]
Rewrite the equation so that the left side is factorable. Add that value to both sides. Use the equation (b/2)2
([latex]\frac{4}{5}[/latex] ÷ 2)2 = [latex]\frac{4}{25}[/latex]
x2 – [latex]\frac{4}{5}[/latex]x + [latex]\frac{4}{25}[/latex] = [latex]\frac{2}{5}[/latex] + [latex]\frac{4}{25}[/latex]
x2 – [latex]\frac{4}{5}[/latex]x + [latex]\frac{4}{25}[/latex] = [latex]\frac{14}{25}[/latex]
Solving for Roots using the Quadratic Formula
[latex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex]
When
b2 – 4ac < 0
there are 2 imaginary roots
b2 – 4ac = 0
there is one real root
b2 – 4ac > 0
there is 2 real roots