Solving Quadratics By Factoring (a≠1) Unit 2 – Day 4

End Warm Up Answer each question to solve x2 – 9x + 18 = 0 Determine the values of a, b and c. Factor the quadratic expression using what you know about b and c. (x )(x ) = 0 Set each factor to zero and solve. 10 minutes End

Homework Check

Today’s Objective Students will continue to solve quadratic equations by factoring but focus on expressions where a ≠ 1.

What happens when a≠1? On Friday, we found a relationship between two binomials and their product in standard form ax2 + bx + c. In each of our examples, a = 1. When a ≠ 1, we first want to see if there is a common factor we can factor out.

Factor 3x2 – 27x + 54 3x2 – 27x + 54 Each term is divisible by 3 3(x2 – 9x + 18) Now a = 1 so we can factor 3(x – 3)(x – 6) Just like in the Warm Up If asked to solve… x – 3 = 0 x – 6 = 0 x = 3 x = 6

Factoring when a ≠ 1 and no common factor When a ≠ 1 and there is no common factor, factoring involves and trial and error process. Factor 3x2 + x – 10 We can’t start with (x )(x ) like before because that won’t give us 3x2. What two values will multiply to 3x2? (3x )(x ) Because c is negative, we need a negative factor and a positive factor. What two numbers will multiply to -10? Plug in your numbers and check them by distributing. If it doesn’t work, try switching the signs or using different factors.

Solving by Factoring when a≠1 Will our solutions just be the opposite of our factors like before? Solve 3x2 + x – 10 = 0 We just factored this into (3x – 5)(x + 2) = 0. Does x = 5 work? (3(5) – 5)(5 + 2) = (10)(7) ≠ 0 NO! We must SOLVE each factor by setting it equal to 0. 3x – 5 = 0 x + 2 = 0 x = 5/3 x = -2

Example 1: Solve 5x2 – 7x – 6 = 0 Remember: Factor by trial and error, then set each factor to 0 Answers: ½, -3/2

Example 2: Solve 10x2 - 22x – 24 = 0 Factor out 2 first! Answers: -4/5, 3

Example 3: Solve 4x2 + 3x = 0 Factor out x first! Answers: 0, -3/4

Homework Complete the worksheet