# Math Help - 4 college algebra questions

1. ## 4 college algebra questions

1. a farmer has 1442 meters of fencing available to enclose a rectangular portion of his land. One side of the rectangle being fenced lies along a river, so only three sides require fencing.
(a) express the area A of the rectangle as a function of x, where x is the
length of the side parallel to the river.
(b) for what value of x is the area largest

2. Use synthetic division to complete the indicated factorization.
x4 -2x3-25x2+26x+120=(x-3)(x+2)( )
A: x2-x-19 B: x2+x+20 C: x2-x-20 D: x2-2x-19

3. Given that one zero of P(x)=x3+7x2-23x-185 is 6-i, which of the following is also a zero of P(x)?
A: 6-I B: -6+I C: 1-6i D: -1-6i

4. Find the complete factorization of 2x4+3x3-12x2-7x+6 if (2x-1) is one of the factors

2. Originally Posted by Lane
1. a farmer has 1442 meters of fencing available to enclose a rectangular portion of his land. One side of the rectangle being fenced lies along a river, so only three sides require fencing.
(a) express the area A of the rectangle as a function of x, where x is the
length of the side parallel to the river.
(b) for what value of x is the area largest
(a) we now the area of a rectangle is length times width, and we know the length is $x$ we also know that both the widths added together are $1442-x$ since there are 2 sides for width we divide what they are added together by 2. $w=\frac{1442-x}{2}$.

so now we take what we know...

$x=\text{length}$
$w=\frac{1442-x}{2}$
and the Area is the function of $x$

and solve for area...
$l\cdot w=A$

$x\cdot\left(\frac{1442-x}{2}\right)=f(x)$

$\frac{1442x-x^2}{2}=f(x)$

$f(x)=\frac{-x^2}{2}+\frac{1442x}{2}$

$f(x)=\frac{-1}{2}x^2+721x$ now we are left with a quadratic equation that opens down, so we find the vertex...

$\text{vertex}=\frac{-b}{2a}$

$\frac{-721}{2\cdot\frac{-1}{2}}$

$\frac{-721}{-1}$

$721$

so the area is largest when $x=721$

3. Originally Posted by Lane

3. Given that one zero of P(x)=x3+7x2-23x-185 is 6-i, which of the following is also a zero of P(x)?
A: 6-I B: -6+I C: 1-6i D: -1-6i
Since, $6-i$ is a solution then $6+i$ is a solution cuz, solutions appear in conjegate paris.

4. Hello, Lane!

2. Use synthetic division to complete the indicated factorization.
. . . $2x^3-25x^2+26x+120 \;= \;(x-3)(x+2)(\qquad)$

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

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: . $x^2 - x - 20$ . . . Answer (C)

4. Find the complete factorization of $2x^4+3x^3-12x^2-7x+6$
if $(2x-1)$ is one of the factors.

Using long division or synthetic division, we find that:
. . $x^4 - 2x^3 - 25x^2 + 26x + 120 \;=\;(2x - 1)(x^3 +2 x^2 - 5x - 6)$

From the Rational Roots theorem, we test: . $x\:=\:\pm1,\;\pm2,\;\pm3$
. . and find that: . $x\:=\:-1,\;2,\;-3$ are zeros of the cubic.

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