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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.

  1. Start with the Standard Form:
    • Ensure the quadratic expression is in the form ax2 + bx + c = 0
  2. Isolate the Quadratic and Linear Terms:
    • If necessary, divide all terms by a so that the coefficient on x2 is equal to 1.
  3. Move the constant term to the right side of the equation
  4. 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
  5. 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

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