# More questions to solve!

• Mar 4th 2010, 01:16 PM
quikwerk
More questions to solve!
I've done my homework this time and I think I have answers to some of my questions!
The thing is I don't know if I'm right.......

Factor the following polynomials.

1
x^2+x-6

I used brute force and got the answer (x+3)(x-2), I do not remember how to show working for this kind of question though.

2
6x^2+13x-5

Same deal here....Brute force, no idea how to show working.....

The area of a right triangle is 250 in^2. find the lengths of its legs if one leg is 5 inches longer than the other.(use the five step method)

I know the lengths: 20 in and 25 in, but as above I use brute force, so--yeah, I'm a little clueless as to how to show my work.

x^3+5x^2=6x

I think that the answer could be 0,1 or -6, I found -6 using a math program so I don't quite know how I would show that--are they all correct, or just 2 of them? Could I just put {-6,0,1} and call that my answer?
• Mar 4th 2010, 01:57 PM
Ben92
1,2:

You have to find the roots of the quadratic equation, because

$\displaystyle 6\left(x-x_{1}\right)\left(x-x_{2}\right)=6x^{2}+13x-5,$

where $\displaystyle x_{1},x_{2}$ are roots of the equation.

3:

It is known that the area of a right triangle is $\displaystyle \frac{ab}{2}$, because that is a half rectangle, where $\displaystyle a,b$ are the legs of the triangle.

$\displaystyle \frac{x(x+5)}{2}=250$
• Mar 4th 2010, 02:22 PM
TwistedOne151
Another method to use when trying to factor a quadratic $\displaystyle ax^2+bx+c$, is to multiply the coefficient of the $\displaystyle x^2$ term and the constant term (ac), then look for a pair of factors for this whose sum is the coefficient of the linear term (b). So for problem 1, ac=1(-6)=-6; so we look for two numbers whose product is -6 and whose sum is 1. Since ac=-6 is negative, one factor will be negative, and the other positive; so our problem is equivalent to finding factors of 6 whose difference is 1. Since the pairs of factors that multiply to 6 are 1 & 6 or 2 & 3, we see that 3-2=1, and so 3*(-2)=-6, and 3+(-2)=1. These two factors form the two x terms produced in FOIL that add to become the x term in the quadratic, so
$\displaystyle x^2+x-6=x^2+3x-2x-6=(x^2+3)+(-2x-6)=x(x+3)-2(x+3)=(x-2)(x+3)$.

Similarly for problem 2, ac=6*(-5)=-30, so we need a pair of numbers that multiply to -30 and add to 13; looking at pairs of factors of 30 and their differences:
30 & 1; difference=29
15 & 2; difference=13*
so we have 15*(-2)=-30, 15+(-2)=13,
so
$\displaystyle 6x^2+13x-5=6x^2+15x-2x-5$
$\displaystyle \;\;\;=(6x^2+15x)+(-2x-5)$
$\displaystyle \;\;\;=3x(2x+5x)-(2x+5)$
$\displaystyle \;\;\;=(3x-1)(2x+5)$.

--Kevin C.
• Mar 5th 2010, 02:04 PM
quikwerk
ROFLMAO!
I forgot to add the answer I found to my second question, I see that I'm a little off with it, but I'll still post it:

2
6x^2+13x-5

(2x-5)(3x+1)
Same deal here....Brute force, no idea how to show working.....

Thank you guys, I think I understand how to work those questions!
WOOOO!!

One small thing:
What do I do here?

x^3+5x^2=6x

I think that the answer could be 0,1 or -6, I found -6 using a math program so I don't quite know how I would show that--are they all correct, or just 2 of them? Could I just put {-6,0,1} and call that my answer?
• Mar 5th 2010, 04:56 PM
Quacky
$\displaystyle x^3+5x^2=6x$
$\displaystyle x^3+5x^2-6x=0$
$\displaystyle x(x^2+5x-6)=0$
For this to be true,
Either $\displaystyle x=0$ (which gives us one solution)
or $\displaystyle x^2+5x-6=0$
Because when timsed, the result must equal 0
By factorizing that quadratic, can you finish from here? (Wink)
• Mar 5th 2010, 05:52 PM
Wilmer
Quote:

Originally Posted by quikwerk
What do I do here?
x^3+5x^2=6x

Set = 0 : x^3 + 5x^2 - 6x = 0
Common factor x: x(x^2 + 5x - 6) = 0
Factor: x(x - 1)(x + 6) = 0
Sooooo: x = 0 or x = 1 or x = -6