Precalculus Skill Builders Vol 1 Solutions

5. Multiplying Binomials
Example:
Multiply (x + 2)(x ! 5) . Each term in the first binomial is distributed over the
second binomial, as shown below.
Solution: (x + 2)(x ! 5) = x(x ! 5) + 2(x ! 5) = x 2 ! 5x + 2x ! 10 = x 2 ! 3x ! 10
Multiply the following binomials.
1. (x + 3)(x ! 6)
[ x 2 ! 3x ! 18 ]
3. (b 2 – 3)(b + 5)
[ b 3 + 5b 2 – 3b – 15 ]
5. (p 11 – p 10 )(p 10 – p 9 )
[ p 21 – 2p 20 + p 19 ]
7. (5h 4 + 1)(6h – 2)
[ 30h 5 – 10h 4 + 6h – 2 ]
9. (x + 3)(x + 6)(x – 7)
[ x 3 + 2x 2 – 45x – 126 ]
11. (m 99 + m 98 )(m 97 – m 96 )
[ m 196 – m 194 ]
13. (100 j 100 – j)(9 j 2 + 1)
{ } ]
[ 900 j 102 + 100 j 100 – 9 j 3 – j
15. (c 2 x + d 3 y)(2x + c 2 d)
[ 2c 2 x 2 + 2d 3 xy + c 4 dx + c 2 d 4 y ]
2. (a 3 + 1)(a 2 – 4)
[ a 5 – 4a 3 + a 2 – 4 ]
4. (z 3 + 1)(z 2 – 1)
[ z 5 – z 3 + z 2 – 1 ]
6. (q 4 + 4)(q 3 + 3)
[ q 7 + 3q 4 + 4q 3 + 12 ]
8. (d 5 – d 4 )(d 6 + 1)
[ d 11 – d 10 + d 5 – d 4 ]
10. (x 2 + 3)(x – 2)(x + 1)
[ x 4 – x 3 + x 2 – 3x – 6 ]
12. (6n 4 + 3n 3 )(6n 3 + 3n 2 )
[ 36n 7 + 36n 6 + 9n 5 ]
14. (2!k 5 + 5k 4 )(!k + 3)
[ 2! 2 k 6 + 11! k 5 + 15k 4 ]
16. ( f 3 y 3 + g 2 x 2 )( fy + gx)
[ f 4 y 4 + f 3 gxy 3 + fg 2 x 2 y + g 3 x 3 ]
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