All About Fractoring Algebraic Equations

**RyanCouture** · June 1st 2007, 07:27 PM

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.

**RyanCouture** · June 1st 2007, 07:55 PM

wow FYI boys check out this site

QuickMath Automatic Math Solutions

**CaptainBlack** · June 1st 2007, 08:54 PM

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

**CaptainBlack** · June 1st 2007, 09:07 PM

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 $m^2+5m+6$ .

Suppose:

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

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

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

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

RonL

**CaptainBlack** · June 1st 2007, 09:46 PM

83 (b) Factor $p^2-pq -12q^2$

Here we are looking for a factorisation of the form:

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

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

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

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

RonL

**CaptainBlack** · June 1st 2007, 10:19 PM

83 (c) Factor $12x^2+19x-18$

Write:

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

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

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

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

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

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

$ 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 $b=-2$ and $a=3$ , and we find that:

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

RonL

**CaptainBlack** · June 1st 2007, 10:26 PM

84 (a) Factor $9y^2-100$

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

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

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

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

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

and trial and error will quickly show that:

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

RonL

**Krizalid** · June 2nd 2007, 08:18 AM

Originally Posted by CaptainBlack

83 (c) Factor $12x^2+19x-18$

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

Originally Posted by CaptainBlack

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

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

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