Question

71 POINTS!!!

Use long division or synthetic division to factor each polynomial completely with the given factor then give the zeros of the function

f(x) = 6x^3+25x^2-29x-20; (2x+1)

Answer 1

Answer:

f(x) = (2x+1)(3x−4)(x+5)

x = -5, x = -1/2, and x = 4/3

Step-by-step explanation:

Start by setting up the long division:

$2x+1)\overline{6x^{3}+25x^{2}-29x-20}$

Look at the first terms.  2x goes into 6x³ a number of 3x² times.

$\phantom{2x+1)6x^{3}+2}3x^{2}\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}$

2x+1 times 3x² is 6x³+3x².

$\phantom{2x+1)6x^{3}+2}3x^{2}\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+1)} 6x^{3}+3x^{2}$

Subtract.

$\phantom{2x+1)6x^{3}+2}3x^{2}\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}$

Now drop down the next term (the -29x).

$\phantom{2x+1)6x^{3}+2}3x^{2}\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}$

Repeat the process.  2x goes into 22x² a number of 11x times.

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}$

Multiply:

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}\\\phantom{2x+1)6x^{3}+}22x^{2}+11x}$

Subtract:

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}\\\phantom{2x+1)6}-(22x^{2}+11x)\\\phantom{2x+1)6x^{3}} \overline{\phantom{+25x^{2}}-40x}$

Drop down the next term (-20):

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}\\\phantom{2x+1)6}-(22x^{2}+11x)\\\phantom{2x+1)6x^{3}} \overline{\phantom{+25x^{2}}-40x}-20$

2x goes into -40x a number of -20 times.

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x-20\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}\\\phantom{2x+1)6}-(22x^{2}+11x)\\\phantom{2x+1)6x^{3}} \overline{\phantom{+25x^{2}}-40x}-20$

Multiply:

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x-20\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}\\\phantom{2x+1)6}-(22x^{2}+11x)\\\phantom{2x+1)6x^{3}} \overline{\phantom{+25x^{2}}-40x}-20\\\phantom{2x+1)6x^{3}+25x} -40x-20$

Subtract:

$\phantom{2x+1)6x^{3}+2}3x^{2}+11x-20\\2x+1)\overline{6x^{3}+25x^{2}-29x-20}\\\phantom{2x+} -(6x^{3}+3x^{2})\\\phantom{2x+1)} \overline{\phantom{6x^{3}+}22x^{2}}-29x}\\\phantom{2x+1)6}-(22x^{2}+11x)\\\phantom{2x+1)6x^{3}} \overline{\phantom{+25x^{2}}-40x}-20\\\phantom{2x+1)6x^{3}+2} -(-40x-20)\\\phantom{2x+1)6x^{3}+25x^{2}} \overline{\phantom{-40x-2}0}$

The remainder is 0.  So f(x) can be factored as:

f(x) = (2x+1)(3x²+11x−20)

Which we can further factor using AC method:

f(x) = (2x+1)(3x−4)(x+5)

So the zeros are x = -5, x = -1/2, and x = 4/3.