SUM OF TWO CUBES:
Name _____________________________
Period _____ Date ________________
SUM OF TWO CUBES:
|PATTERN: | |a 3 |+ |b 3 |= ( |a |
| | | | | |= ( |___ |
| | | |
|x 2 (x – 2) |-9 (x – 2) |Take out the GCF of the first two, and the GCF of the last two. |
| | |If the first term is negative, FACTOR OUT A NEGATIVE SIGN. |
|(x 2 – 9)(x – 2) = 0 |Factor out the common factor—the “leftovers” that match. |
| |For example: x 2 (() – 9(() = (x 2 – 9)( |
| |(Factor out the () |
|(x + 3)(x – 3)(x – 2) = 0 |Keep factoring! (x 2 – 9) = (x + 3)(x – 3) |
| |(Difference of perfect squares) |
|x = 3, -3, 2 |Set each factor equal to 0 and solve. |
ANOTHER EXAMPLE:
Solve 2b 4 + 14b 3 = 16b + 112
|2b 5 + 14b 4 – 16b 2 – 112b = 0 |Get 0 on one side. |
|2b(b 4 + 7b 3 – 8b – 56) = 0 |Factor out the GCF. Keep factoring the rest! |
|b 4 + 7b 3 |-8b – 56 |GCF of first pair; GCF of last pair. |
|b 3(b + 7) |-8(b + 7) | |
|2b(b 3 – 8)(b + 7) = 0 |Combine. Remember your GCF! |
|2b(b – 2)(b 2 + 2b + 4)(b + 7) = 0 |Factor b 3 – 8 further. |
|2b = 0 ( b = 0 |Set each factor equal to 0 and solve |
|b – 2 = 0 ( b = 2 | |
|b + 7 = 0 ( b = -7 | |
|b 2 + 2b + 4 = 0 ( not factorable ( | |
|So use the quadratic formula: | |
| [pic] |
|Solutions are b = 0, 2, -7, and [pic] |
PRACTICE:
Solve each equation. SHOW YOUR WORK ON A SEPARATE SHEET OF PAPER.
|6. |r 3 – 3r 2 + 6r = 18 | |7. |x 3 + 6x 2 + 7x + 42 = 0 |
|8. |c 3 + 4c 2 – 9c – 36 = 0 | |9. |9m 4 + 18m 3 – 4m 2 – 8m = 0 |
|10. |x 3 + 2x 2 = 25x + 50 | |11. |6(w 3 + 5w 2 – 2w) = 6w + 90 |
|12. |6(r 7 + r 5) = 9(r 6 + r 4) | |13. |x 3 = 2(x 2 + 8x – 16) |
|14. |16d 3(d + 2) = 2d + 4 | |15. |125a 4 – 27 = 125a 3 – 27a |
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