Lesson · Grade 11 · Pre Calculus

Factoring by Substitution

  • Grade 11
  • Pre Calculus
  • 1 min 58 s
  • British Columbia
  • BC11-MA-112

The video factors 2(x + 1)² + 5(x + 1) − 3 by letting u = x + 1, factoring 2u² + 5u − 3 into (2u − 1)(u + 3), and then putting x + 1 back to get (2x + 1)(x + 4). It warns against stopping at the u form, checks the answer by expanding both sides to 2x² + 9x + 4, and ends on a three-step recap.

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In one sentence

The question this lesson answers

Let us factor this expression, which means writing it as a product of brackets.

Notice that the same bracket, x plus 1, appears twice.

Transcript

What the lesson says

Let us factor this expression, which means writing it as a product of brackets.

Notice that the same bracket, x plus 1, appears twice.

So we use substitution: replace the repeated bracket with one new letter. Let u equal x plus 1.

Each x plus 1 becomes u: 2 u squared plus 5 u minus 3. That is a quadratic in u, since its highest power is u squared.

Factor it the usual way. 2 times negative 3 is negative 6, so find two numbers that multiply to negative 6 and add to 5.

6 and negative 1 work: 6 times negative 1 is negative 6, and 6 plus negative 1 is 5.

Split 5 u into 6 u minus u, then take a common factor out of each pair.

Both parts share u plus 3, so we get 2 u minus 1, times u plus 3.

A common mistake is to stop here, but this is still in u. Put x plus 1 back in place of each u.

Remove the inner brackets: 2 times x plus 1 is 2 x plus 2.

Then 2 minus 1 is 1, and 1 plus 3 is 4. So the answer is 2 x plus 1, times x plus 4.

Check by expanding, which means multiplying out. Our answer gives 2 x squared plus 9 x plus 4.

The start also expands to 2 x squared plus 9 x plus 4. They match, so the factoring is correct.

To recap: let u stand for the repeated bracket, factor in u, then put x plus 1 back.

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