# Thread: Solving Polynomial Equations

1. ## Solving Polynomial Equations

I have a math assignment due tomorrow and there are a few questions that I don't even know where to begin, how to do, etc. Just overall stumpers. A group of 8 people from the class have been involved in an msn convo for the last 2hr trying to figure them out.. with no luck whatsoever. Any help/tips/start offs, etc. would be very much appreciated!

1) (3x-5)(3x+1)² (3x+7) + 68 = 0

-1
± √
34 , -1 ± √ 2
..................... 3 ............. 3 ....

2) (x
²+6x+6)(x²+6x+8) = 528

-8, 2, - 3
± i 21

3) The height, length, and width of a small box are consecutive integers with the height being the smallest of the three dimensions. If the length and width are increased by 1cm each and the height is doubled, then the volume is increased by 120cm
³.
Find the dimensions of the original box.

Answer: 3cm, 4cm, 5cm

2. Originally Posted by AnnieD
I have a math assignment due tomorrow and there are a few questions that I don't even know where to begin, how to do, etc. Just overall stumpers. A group of 8 people from the class have been involved in an msn convo for the last 2hr trying to figure them out.. with no luck whatsoever. Any help/tips/start offs, etc. would be very much appreciated!

1) (3x-5)(3x+1)² (3x+7) + 68 = 0

-1
± √
34 , -1 ± √ 2
..................... 3 ............. 3 ....

ok, so i can't see an easy way to do this one. hopefully we'll get back to it

you can use the quartic formula, but that is really a last resort, it's messy

2) (x²+6x+6)(x²+6x+8) = 528

-8, 2, - 3
± i 21

you need to know either long division of polynomials or synthetic division to do this question.

We have $\left( x^2 + 6x + 6 \right) \left( x^2 + 6x + 8 \right) = 528$

expanding the brackets and subtracting 528 from both sides, we get:

$x^4 + 12x^3 + 50x^2 + 84x - 480 = 0$

trying the factors of 480, we find that $x = 2$ is a root of the above quartic, thus, by the factor theorem, $(x - 2)$ is a factor.

performing polynomial long division/synthetic division on the above, we find that:

$x^4 + 12x^3 + 50x^2 + 84x - 480 = (x - 2) \left( x^3 + 14x^2 + 78x + 240 \right) = 0$

trying the factors of 240, we find that $x = -8$ is a root of the above cubic, thus by the factor theorem, $(x + 8)$ is a factor.

performing polynomial long division/synthetic division on the above, we find that:

$(x - 2) \left( x^3 + 14x^2 + 78x + 240 \right) = (x - 2)(x + 8) \left( x^2 + 6x + 30 \right) = 0$

$\Rightarrow x - 2 = 0 \mbox { or } x + 8 = 0 \mbox { or } x^2 + 6x + 30 = 0$

For the first two equations, obviously $x = 2$ and $x = -8$ are the roots. for the quadratic, we use the quadratic formula and simplify to obtain the two remaining roots, which are $x = -3 \pm \sqrt {21}~i$

3) The height, length, and width of a small box are consecutive integers with the height being the smallest of the three dimensions. If the length and width are increased by 1cm each and the height is doubled, then the volume is increased by 120cm³.
Find the dimensions of the original box.

Answer: 3cm, 4cm, 5cm
Let the height be $n$
Let the width be $n + 1$
Let the length be $n + 2$

the volume of this box is given by:

$V = n(n + 1)(n + 2)$

if the length and width are increased by 1 and the height is doubled, the volume, V, increases by 120.

This means that $V + 120 = {\color {red} 2}n(n + 1 {\color {red} + 1})(n + 2 {\color {red} + 1})$

$\Rightarrow V = 2n(n + 2)(n + 3) - 120$

but $V = n(n + 1)(n + 2)$

so we must have:

$2n(n + 2)(n + 3) - 120 = n(n + 1)(n + 2)$

solve the above for $n$ and you can find the dimensions of the original box

3. Hello, Annie!

$1)\;\;(3x-5)(3x+1)^2(3x+7) + 68 \:=\: 0$

Answers: . $\frac{-1 \pm \sqrt{34}}{3},\;\;\frac{-1 \pm \sqrt{2}}{3}$

Let: . $u \,= \,3x + 1$

Then we have: . $(u-6)u^2(u+6) + 68 \:=\:0\quad\Rightarrow\quad u^4-36u^2 + 68 \:=\:0$

. . which factors: . $(u^2-34)(u^2-2) \:=\:0$

. . and has roots: . $u \:=\:\pm\sqrt{34},\:\pm\sqrt{2}$

Back-substitute: . $\begin{Bmatrix}3x+1\:=\:\pm\sqrt{34} & \Rightarrow & x \:=\:\boxed{\frac{-1\pm\sqrt{34}}{3}} \\ \\
3x+1 \:=\:\pm\sqrt{2} & \Rightarrow & x \:=\:\boxed{\frac{-1\pm\sqrt{2}}{3}} \end{Bmatrix}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can use a similar trick on #2 . . .

$2)\;\;(x^2+6x+6)(x^2+6x+8) \:=\:528$

Let: . $u \:=\:x^2+6x+7$

Then we have: . $(u-1)(u+1) \:=\:528\quad\Rightarrow\quad u^2-529\:=\:0$

. . which factors: . $(u - 23)(u + 23) \:=\:0$

. . and has roots: . $u \:=\:\pm23$

Back-substitute:

$x^2+6x+7\:=\:23\quad\Rightarrow\quad x^2+6x-16\:=\:0$

. . $(x + 8)(x - 2)\:=\:0 \quad\Rightarrow\quad x \:=\:\boxed{-8,\:2}$

$x^2+6x+7\:=\:-23\quad\Rightarrow\quad x^2 + 6x + 30 \:=\:0$

. . $x \:=\:\frac{-6\pm\sqrt{-84}}{2} \;=\;\boxed{-3\pm i\sqrt{21}}$

4. Thank you both oh so very very much!!

5. Originally Posted by Soroban
Hello, Annie!

Let: . $u \,= \,3x + 1$

Then we have: . $(u-6)u^2(u+6) + 68 \:=\:0\quad\Rightarrow\quad u^4-36u^2 + 68 \:=\:0$

. . which factors: . $(u^2-34)(u^2-2) \:=\:0$

. . and has roots: . $u \:=\:\pm\sqrt{34},\:\pm\sqrt{2}$

Back-substitute: . $\begin{Bmatrix}3x+1\:=\:\pm\sqrt{34} & \Rightarrow & x \:=\:\boxed{\frac{-1\pm\sqrt{34}}{3}} \\ \\
3x+1 \:=\:\pm\sqrt{2} & \Rightarrow & x \:=\:\boxed{\frac{-1\pm\sqrt{2}}{3}} \end{Bmatrix}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can use a similar trick on #2 . . .

Let: . $u \:=\:x^2+6x+7$

Then we have: . $(u-1)(u+1) \:=\:528\quad\Rightarrow\quad u^2-529\:=\:0$

. . which factors: . $(u - 23)(u + 23) \:=\:0$

. . and has roots: . $u \:=\:\pm23$

Back-substitute:

$x^2+6x+7\:=\:23\quad\Rightarrow\quad x^2+6x-16\:=\:0$

. . $(x + 8)(x - 2)\:=\:0 \quad\Rightarrow\quad x \:=\:\boxed{-8,\:2}$

$x^2+6x+7\:=\:-23\quad\Rightarrow\quad x^2 + 6x + 30 \:=\:0$

. . $x \:=\:\frac{-6\pm\sqrt{-84}}{2} \;=\;\boxed{-3\pm i\sqrt{21}}$

Genius, i tell you! Pure Genius! You rock, Soroban!!!

i particularly liked your cleaver substitution for the second one. Doing it in such a way that you obtain the difference of 2 squares saves you a LOT of trouble. Amazing!

I'm going to file away this trick and use it later, save myself the trouble of having to simplify these quartics