Pre Calculus 11 Week 10: Reviewing Factoring with Substitution

Last week in Pre Calculus 11 we mostly reviewed for our midterm exam. As I was studying I came across a few things that I totally forgot about. Things that would help make my math life easier, like how to use substitution when factoring. This is helpful because it is simple, faster, and not as messy as factoring without this trick. I will quickly review how to do this.

Example: 3(2x-1)^2 + 14(2x-1) + 8

Step 1: Substitute (2x-1) with a variable.

\star Notice that there are two (2x-1), therefore you can use the same variable for both terms.

  • 3a^2 + 14a + 8

Step 2: Factor3a^2 + 14a + 8

  • (3a + 2) (a + 4)

Step 3: Substitute (2x – 1) back into the factored form to replace the variable a.

  • (3a + 2) (a + 4)
  • (3(2x -1) + 2) ((2x-1)+ 4

Step 4: Distribute

  • (3(2x -1) + 2) ((2x-1)+ 4
  • (6x-3 +2)(2x-1+4)

Step 5: Simplify

  • (6x-3 +2)(2x-1+4)
  • (6x-1)(2x+3)

FINAL ANSWER

(6x-1)(2x+3)

To find the solutions make each factor equal to zero and solve for x:

6x-1=0 –> 6x=1 –> x=\frac{1}{6}

2x + 3=0 –>2x=-3 –> x =- \frac{3}{2}

This is how to use substitution while factoring. This is a very helpful and quick way to factor difficult looking equations.

 

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