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)

The numbers at the bottom stand for the coefficients of the answer, which is .
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.

The remainder is 1746, so P(5)=1746. / 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.

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) = 2x3 + 7x2 – 53x – 28; and 4 is a zero.