Worksheet 5.4 & 5.5
Algebra 2
Factoring: Look for a greatest common
factor and then look at the number of terms
of the problem for a clue on how to start
the factoring. See the notes on factoring
(before Ch. 5) for more examples.
Synthetic Division: Divide using coefficients.
Ex.: Simplify (x3 – 7x – 6) ÷ (x – 2)
2 1 0 7 6
2
4
6
1 2  3  12
Remainder Theorem: The remainder
from synthetic division is the value of the
function at that point.
Ex: Use synthetic division to find the
value of P(5) if P(x) = 3x4 – 5x2 + x – 9.
The value for x is 5 because of the
function notation. Put 5 outside the box
and put the coefficients of the polynomial
inside the box. Remember to use all
descending powers, so there’s a 0x3.
0
5
15
75 350 1755
5 3
1
9
3 15 70 351 1746
The remainder is 1746, so P(5)=1746.
The numbers at the bottom stand for the coefficients of the
answer, which is x 2  2 x  3  x122 .
Factor Theorem: If a remainder is zero, the number outside the
box in synthetic division is a zero of the function; therefore, the
related factor is a factor of the polynomial.
Ex: Factor f(x) = 2x3 + 11x2 + 18x + 9, given that f(-3) = 0.
The value for x is -3 because of the f(x) notation. Put -3 outside
the box, and the coefficients of the polynomial inside the box.
3 2
2
11 18
9
 6  15  9
5
3
0
The remainder is 0, so x = -3 is a zero of the function. The
remainder of the problem is 2x2 + 5x + 3, which factors as (2x +
3)(x + 1). x = -3 becomes the factor (x + 3) when you move the -3
over.
The factors of 2x3 + 11x2 + 18x + 9 are (2x + 3),(x + 1),
and (x + 3).
Factor.
1. 216x3 + 1
2. 3x2 + 11x + 6
3. x3 – 4x2 + 4x – 16
4. x4 + 3x2 + 2
5. 2x7 – 32x3
6. 4x4 – 5x2 – 9
Solve by factoring. (Find real and/or imaginary answers.)
7. x3 + 2x2 – x = 2
8. x3 + 8 = 0
9. 3x4 + 15x2 – 72 = 0
Divide using synthetic division.
10. ( x3 – 14x + 8) ÷ (x + 4)
11. (x4 – 6x3 – 40x + 33) ÷ (x – 7) 12. (3x2 – 10x) ÷ (x – 6)
Factor the polynomial.
13. f(x) = x3 – 3x2 – 16x – 12, and f(6) = 0.
14. f(x) = x3 – 11x2 + 14x + 80, and f(8) = 0.
15. f(x) = 2x3 + 7x2 – 33x – 18, and f(-6) = 0.
16. f(x) = x3 – x2 – 21x + 45, and f(-5) = 0.
Find the zeros of the function given that one of the zeros is ___.
17. f(x) = 2x3 + 3x2 – 39x – 20, and 4 is a zero.
18. f(x) = x3 + 11x2 – 150x – 1512, & -14 is a zero.
19. f(x) = x3 + x2 +2x + 24, and -3 is a zero.
20. f(x) = 4x3 + 9x2 – 52x + 15, and -5 is a zero.