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EXAMPLE 1

EXAMPLE 1. Factor ax 2 + bx + c where c > 0. Factor 5 x 2 – 17 x + 6. SOLUTION.

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EXAMPLE 1

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  1. EXAMPLE 1 Factor ax2 + bx + c where c > 0 Factor 5x2 – 17x + 6. SOLUTION You want 5x2 – 17x + 6 = (kx + m) (lx + n) where kand lare factors of 5 andmand nare factors of 6. You can assume that kand lare positive and k ≥ l. Because mn> 0, mand nhave the same sign. So, mand nmust both be negative because the coefficient of x, – 17, is negative.

  2. ANSWER The correct factorization is5x2 – 17x + 6 = (5x – 2) (x – 3). EXAMPLE 1 Factor ax2 + bx + c where c > 0

  3. ANSWER The correct factorization is3x2 + 20x – 7= (3x – 1) (x + 7). EXAMPLE 2 Factor ax2 +bx + c where c < 0 Factor 3x2 + 20x – 7. SOLUTION You want3x2 + 20x – 7 = (kx + m) (lx + n)wherekandlarefactors of3andmandnare factors of– 7. Becausemn < 0, mandn have opposite signs.

  4. for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 1. 7x2 – 20x – 3 SOLUTION You want 7x2 – 20x – 3 = (kx + m) (lx + n) where kand lare factors of 7 andmand nare factors of 3. You can assume that kand lare positive and k ≥ l. Because mn< 0, mand nhave opposite signs.

  5. ANSWER The correct factorization is7x2 – 20x – 3= (7x + 1) (x – 3). for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE

  6. for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE 2. 5z2 + 16z + 3 SOLUTION You want 5z2 + 16z + 3 = (kx + m) (lx + n) where kand lare factors of 5 andmandnare factors of3. You can assume that kand lare positive and k ≥ l. Because mn> 0, mand nhave the same sign.

  7. ANSWER The correct factorization is5z2 + 16z + 3 = (5z + 1) (z + 3). for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE

  8. for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE 3. 2w2 + w + 3 SOLUTION You want 2w2 + w + 3 = (kx + m) (lx + n) where kand lare factors of 2 andmand nare factors of 3. Because mn> 0, mand nhave the same sign.

  9. ANSWER 2w2 + w + 3 cannot be factored for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE

  10. for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE 4. 3x2 + 5x – 12 SOLUTION You want 3x2 + 5x – 12 = (kx + m) (lx + n) where kand lare factors of 3 andmand nare factors of – 12 . Because mn< 0, mand nhave opposite sign.

  11. ANSWER The correct factorization is3x2 + 5x – 12 = (3x – 4) (x + 3). for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE

  12. for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE 5. 4u2 + 12u + 5 SOLUTION You want 4u2 + 12u + 5 = (kx + m) (lx + n) where kand lare factors of 4 andmand nare factors of 5. You can assume that kand lare positive and k ≥ l. Because mn> 0, mand nhave the same sign.

  13. ANSWER The correct factorization is4u2 + 12u + 5 = (2u + 1) (2u + 5). for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE

  14. for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE 6. 4x2 – 9x + 2 SOLUTION You want 4x2 – 9x + 2 = (kx + m) (lx + n) where kand lare factors of 4 andmand nare factors of 2. You can assume that kand lare positive and k ≥ l. Because mn> 0, mand nhave the same sign. So, mand nmust both be negative because the coefficient of x =– 9, is negative.

  15. ANSWER The correct factorization is4x2 – 9x + 2 = (4x – 1) (x - 2). for Examples 1 and 2 GUIDED PRACTICE GUIDED PRACTICE

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