Splash Screen Solving x² + bx + c = 0 Lesson 8-6.

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Splash Screen Solving x² + bx + c = 0 Lesson 8-6

Over Lesson 8–5

Then/Now Understand how to factor trinomials and solve equations of the form x 2 + bx + c = 0.

Vocabulary

Concept

Example 1 b and c are Positive Factor x 2 + 7x In this trinomial, b = 7 and c = 12. You need to find two positive factors with a sum of 7 and a product of 12. Make an organized list of the factors of 12, and look for the pair of factors with a sum of 7. 1, , 6 8 3, 4 7The correct factors are 3 and 4. Factors of 12 Sum of Factors

Example 1 b and c are Positive = (x + 3)(x + 4)m = 3 and p = 4 CheckYou can check the result by multiplying the two factors. F O I L (x + 3)(x + 4) = x 2 + 4x + 3x + 12FOIL method = x 2 + 7x + 12Simplify. Answer: (x + 3)(x + 4) x 2 + 7x + 12 = (x + m)(x + p)Write the pattern.

Example 1 Factor x 2 + 3x + 2.

Example 2 b is Negative and c is Positive Factor x 2 – 12x In this trinomial, b = –12 and c = 27. This means m + p is negative and mp is positive. So, m and p must both be negative. Make a list of the negative factors of 27, and look for the pair with a sum of –12. –1,–27–28 –3,–9–12The correct factors are –3 and –9. Factors of 27 Sum of Factors

Example 2 b is Negative and c is Positive = (x – 3)(x – 9)m = –3 and p = –9 Answer: (x – 3)(x – 9) x 2 – 12x + 27 = (x + m)(x + p)Write the pattern.

Example 2 Factor x 2 – 10x + 16.

Example 3 c is Negative A. Factor x 2 + 3x – 18. In this trinomial, b = 3 and c = –18. This means m + p is positive and mp is negative, so either m or p is negative, but not both. Therefore, make a list of the factors of –18 where one factor of each pair is negative. Look for the pair of factors with a sum of 3.

Example 3 c is Negative 1,–18–17 –1, ,–9 –7 –2,9 7 3,–6 –3 –3,6 3The correct factors are –3 and 6. Factors of –18 Sum of Factors

Example 3 c is Negative x 2 + 3x – 18= (x + m)(x + p)Write the pattern. = (x – 3)(x + 6)m = –3 and p = 6 Answer: (x – 3)(x + 6)

Example 3 c is Negative B. Factor x 2 – x – 20. Since b = –1 and c = –20, m + p is negative and mp is negative. So either m or p is negative, but not both. 1,–20–19 –1, ,–10 –8 –2,10 8 4,–5 –1 –4,5 1The correct factors are 4 and –5. Factors of –20 Sum of Factors

Example 3 c is Negative = (x + 4)(x – 5)m = 4 and p = –5 x 2 – x – 20 = (x + m)(x + p)Write the pattern. Answer: (x + 4)(x – 5)

Example 3 A. Factor x 2 + 4x – 5.

Example 3 B. Factor x 2 – 5x – 24.

Example 4 Solve an Equation by Factoring Solve x 2 + 2x = 15. Check your solution.

Example 4 Solve an Equation by Factoring Check Substitute –5 and 3 for x in the original equation. x 2 + 2x – 15= 0x 2 + 2x – 15=0 ? ? (–5) 2 + 2(–5) – 15 = (3) – 15 = 0 0 = 0 0 = 0 ? ? 25 + (–10) – 15 = – 15 = 0

Example 4 Solve x 2 – 20 = x. Check your solution.

Example 5 Solve a Problem by Factoring ARCHITECTURE Marion wants to build a new art studio that has three times the area of her old studio by increasing the length and width by the same amount. What should be the dimensions of the new studio? UnderstandYou want to find the length and width of the new studio.

Example 5 Solve a Problem by Factoring PlanLet x = the amount added to each dimension of the studio. The new length times the new width equals the new area. x + 12 ● x + 10 = 3(12)(10) old area Solve(x + 12)(x + 10) = 3(12)(10)Write the equation. x x = 360Multiply. x x – 240 = 0Subtract 360 from each side.

Example 5 Solve a Problem by Factoring (x + 30)(x – 8)=0Factor. Answer: The length of the new studio should be or 20 feet, and the new width should be or 18 feet. x + 30=0 or x – 8=0Zero Product Property x= –30x=8Solve each equation. Since dimensions cannot be negative, the amount added to each dimension is 8 feet.

Example 5 Solve a Problem by Factoring CheckThe area of the old studio was 12 ● 10 or 120 square feet. The area of the new studio is 18 ● 20 or 360 square feet, which is three times the area of the old studio.

Example 5 A.6 × –8 B.6 × 8 C.8 × 12 D.12 × 18 PHOTOGRAPHY Adina has a 4 × 6 photograph. She wants to enlarge the photograph by increasing the length and width by the same amount. What dimensions of the enlarged photograph will produce an area twice the area of the original photograph?

End of the Lesson Homework p. 507 #12-45 odd