Hello, Lane!

2. Use synthetic division to complete the indicated factorization.

. . . $\displaystyle 2x^3-25x^2+26x+120 \;= \;(x-3)(x+2)(\qquad)$

. . . $\displaystyle A)\;x^2-x-19\qquad B)\; x^2+x+20\qquad C)$$\displaystyle x^2-x-20\qquad D)\;x^2-2x-19$

Do you how to do synthetic division? Code:

3 | 1 -2 -25 26 120
| 3 3 -66 -120
| - - - - - - - - - - - - -
1 1 -22 -40 0
-2 | 1 1 -22 -40
| -2 2 40
| - - - - - - - - - -
1 -1 -20 0

The third factor is: .$\displaystyle x^2 - x - 20$ . . . Answer (C)

4. Find the complete factorization of $\displaystyle 2x^4+3x^3-12x^2-7x+6$

if $\displaystyle (2x-1)$ is one of the factors.

Using long division or synthetic division, we find that:

. . $\displaystyle x^4 - 2x^3 - 25x^2 + 26x + 120 \;=\;(2x - 1)(x^3 +2 x^2 - 5x - 6)$

From the Rational Roots theorem, we test: .$\displaystyle x\:=\:\pm1,\;\pm2,\;\pm3$

. . and find that: .$\displaystyle x\:=\:-1,\;2,\;-3$ are zeros of the cubic.

Therefore, the factorization is: .$\displaystyle (2x - 1)(x + 1)(x - 2)(x + 3)$