# All About Fractoring Algebraic Equations

• Jun 1st 2007, 07:27 PM
RyanCouture
Hey guys, first post here (hopefully the last) - in a good way

Anyway, I'm doing a mail-in correspondance course for grade 12 college math, and I have to have this thing completed within the next day. I breezed through 19 of 20 lessons, and now my arch enemy has returned.

I don't know if its so much that I can't understand it, as it is just very poorly demonstrated and explained in the book here (having no teacher certainly makes things harder). Anyway, I got about 9 questions here that I need you folks to help me out with. I'm just gonna scan em' right from the book.

http://i37.photobucket.com/albums/e9...ame32/math.jpg

There you go, I need help with #83-#85. Any help is welcomed.

Thanks guys.
• Jun 1st 2007, 07:55 PM
RyanCouture
wow FYI boys check out this site

QuickMath Automatic Math Solutions
• Jun 1st 2007, 08:54 PM
CaptainBlack
Quote:

Originally Posted by RyanCouture
wow FYI boys check out this site

QuickMath Automatic Math Solutions

1. We know

2. It does not tell you why, which is the purpose of maths education.

RonL
• Jun 1st 2007, 09:07 PM
CaptainBlack
Quote:

Originally Posted by RyanCouture
Hey guys, first post here (hopefully the last) - in a good way

Anyway, I'm doing a mail-in correspondance course for grade 12 college math, and I have to have this thing completed within the next day. I breezed through 19 of 20 lessons, and now my arch enemy has returned.

I don't know if its so much that I can't understand it, as it is just very poorly demonstrated and explained in the book here (having no teacher certainly makes things harder). Anyway, I got about 9 questions here that I need you folks to help me out with. I'm just gonna scan em' right from the book.

There you go, I need help with #83-#85. Any help is welcomed.

Thanks guys.

83 (a) Factor $\displaystyle m^2+5m+6$.

Suppose:

$\displaystyle m^2+5m+6=(m+a)(m+b)$

Then $\displaystyle a+b=5$ and $\displaystyle ab=6$.

So we try the factors of $\displaystyle 6$ as possible values of $\displaystyle a$ and $\displaystyle b$, and find that $\displaystyle a=2$ and $\displaystyle b=3$ will do, so:

$\displaystyle m^2+5m+6=(x+2)(x+3)$.

RonL
• Jun 1st 2007, 09:46 PM
CaptainBlack
83 (b) Factor $\displaystyle p^2-pq -12q^2$

Here we are looking for a factorisation of the form:

$\displaystyle p^2-pq -12q^2=(p+aq)(p+bq)$,

so $\displaystyle ab=-12$, and $\displaystyle a+b=-1$.

Now we try the factors of $\displaystyle -12$ for $\displaystyle a$ and $\displaystyle b$. These are $\displaystyle 1,\ -1,\ 12,\ -12,\ 2,\ -2,\ 3,\ -3,\ 4,\ -4,\ 6,\ -6$. Of these $\displaystyle -4$ and $\displaystyle 3$ seem to work, and so:

$\displaystyle p^2-pq -12q^2=(p-4q)(p+3q)$

RonL
• Jun 1st 2007, 10:19 PM
CaptainBlack
83 (c) Factor $\displaystyle 12x^2+19x-18$

Write:

$\displaystyle 12x^2+19x-18=(ax+b)(cx+d)$

Then $\displaystyle -18=bd$ and $\displaystyle 12=ac$, so we try the factors
of $\displaystyle -18$ for $\displaystyle b$ and $\displaystyle d$ and the factors of $\displaystyle 12$ for $\displaystyle a$ and $\displaystyle c$.

If you run through all of these you wil be here all day.

A short cut is to use the quadraic formula since if:

$\displaystyle 12x^2+19x-18=(ax+b)(cx+d)$

then the roots of $\displaystyle 12x^2+19x-18=0$ are $\displaystyle -d/c$ and $\displaystyle -b/a$, but the roots are:

$\displaystyle x=\frac{-19 \pm \sqrt{19^2+4\times 12 \times 18}}{2\times 12}=\frac{-9}{4} \mbox{ or } \frac{2}{3}$

So we try $\displaystyle b=-2$ and $\displaystyle a=3$, and we find that:

$\displaystyle 12x^2+19x-18=(3x-2)(4x+9)$

RonL
• Jun 1st 2007, 10:26 PM
CaptainBlack
84 (a) Factor $\displaystyle 9y^2-100$

This is the difference of two squares and you should know the factorisation
of such expressions:

$\displaystyle (a^2-b^2)=(a+b)(a-b)$

84 (b) Factor $\displaystyle 7p^2+19p-6$

is like 83 (c), but as $\displaystyle 7$ is prime we know if this has a "nice" factorisation it is of the form:

$\displaystyle 7p^2+19p-6=(7p+a)(p+b)$

and trial and error will quickly show that:

$\displaystyle 7p^2+19p-6=(7p-2)(p+3)$

RonL
• Jun 2nd 2007, 08:18 AM
Krizalid
Quote:

Originally Posted by CaptainBlack
83 (c) Factor $\displaystyle 12x^2+19x-18$

\displaystyle \begin{aligned}12x^2+19x-18&=&12x^2+(27x-8x)-18\\&=&(12x^2-8x)+(27x-18)\\&=&4x(3x-2)+9(3x-2)\\&=&(3x-2)(4x+9)\end{aligned}

Quote:

Originally Posted by CaptainBlack
84 (b) Factor $\displaystyle 7p^2+19p-6$

\displaystyle \begin{aligned}7p^2+19p-6&=&7p^2+(21p-2p)-6\\&=&7p(p+3)-2(p+3)\\&=&(p+3)(7p-2)\end{aligned}

Another ways :D