Hi all,
First post as you can see. I'm doing high school maths and really need to improve. I can do every exercise in the basic algebra chapter except for these 2. I have no idea why, they've just stumped me.
So any help would be much appreciated.
Hi all,
First post as you can see. I'm doing high school maths and really need to improve. I can do every exercise in the basic algebra chapter except for these 2. I have no idea why, they've just stumped me.
So any help would be much appreciated.
$\displaystyle \frac {8q^2 + 2qr - 6r^2 + 8qs - 6rs}{8q^2 - 2qr - 3r^2 + 12 qs - 9rs}$
$\displaystyle = \frac {2 \left( 4q^2 + qr - 3r^2 + 4qs - 3rs \right)}{(2q + r)(4q - 3r) + 3s(4q - 3r)}$
$\displaystyle = \frac {2 [(4q - 3r)(q + r) + s(4q - 3r)]}{(4q - 3r)(2q + r + 3s)}$
$\displaystyle = \frac {2(4q - 3r)(q + r + s)}{(4q - 3r)(2q + r + 3s)}$
$\displaystyle = \frac {2(q + r + s)}{2q + r + 3s}$
$\displaystyle \frac {15g^2 k^3 + 35g^3 k^2 - 30g^4 k}{8gk^4 + 44g^2 k^3 + 60g^3 k^2}$
$\displaystyle = \frac {5g^2 k \left( 3k^2 + 7gk - 6g^2 \right)}{4g k^2 \left( 2k^2 + 11gk + 15g^2\right)}$
$\displaystyle = \frac {5g^2 k (3k - 2g)(k + 3g)}{4g k^2 (2k + 5g)(k + 3g)}$
$\displaystyle = \frac {5g(3k - 2g)}{4k(2k + 5g)}$
Hello,
to #1:
You have to use the method here that you add a sum with the value zero to the numerator and to the denominator. I'll put the two summands in question into boxes:
$\displaystyle \frac{8q^2+2qr + \boxed{6qr} - \boxed{6qr}-6r^2+8qs - 6rs}{8q^2-2qr -\boxed{4qr}+\boxed{4qr}-3r^2+12qs-9rs} = \text{ } $Now re-arrange numerator and denomintor:
$\displaystyle \frac{8q^2+8qr + 8qs-6qr-6r^2 - 6rs}{8q^2-6qr +4qr-3r^2+12qs-9rs} =\ \ \ $ Now factorize numerator and denominator:
$\displaystyle \frac{8q(q+r + s)-6r(q+r+s)}{2q(4q-3r) +r(4q-3r)+3s(4q-3r)} =\ \ \ $ Now factor again and cancel the common factor which is: (4q - 3r)
I'll leave the rest for you.
Hello, lucius!
Welcome aboard!
There may be two reasons for your difficulty:Factor: .$\displaystyle 2m^2 + 3mn + n^2$
. . [1] There are two variables
. . [2] That "2" in front.
Just as we "split" the first term into two factors,
. . we can do the same with the last term.
It looks like this: .$\displaystyle (\:m\qquad n)(\:m \qquad n)$
Since $\displaystyle 2m^2$ factors into $\displaystyle (2m)(m)$ .and $\displaystyle n^2$ factors into $\displaystyle (n)(n)$,
. . we have: .$\displaystyle (2m\qquad n)(m\qquad n)$
Now insert the signs that make the "middle term" come out to $\displaystyle 3mn$.
Answer: .$\displaystyle (2m + n)(m + n)$
I'll take baby-steps . . .Simplify: .$\displaystyle \frac{1}{c+2} + \frac{2}{c-2} + \frac{4}{c^2-4}$
Factoring the denominators, we have: .$\displaystyle \frac{1}{c+2} + \frac{2}{c-2} + \frac{4}{(c+2)(c-2)}$
To add or subtract fractions, we must have a common denominator.
. . In this problem, it is: $\displaystyle (c+2)(c-2)$
Now we must "convert" each fraction so it has $\displaystyle (c+2)(c-2)$ in the denominator.
Multiply the first fraction by .$\displaystyle \frac{c-2}{c-2}$
Multiply the second fraction by .$\displaystyle \frac{c+2}{c+2}$
The problem looks like this: .$\displaystyle \frac{c-2}{c-2}\cdot\frac{1}{c+2} \:+\: \frac{c+2}{c+2}\cdot\frac{2}{c-2} \:+\: \frac{4}{(c+2)(c-2)} $
. . and we have: .$\displaystyle \frac{c-2}{(c+2)(c-2)} + \frac{2(c+2)}{(c+2)(c-2)} + \frac{4}{(c+2)(c-2)} $
The fractions have a common denominator, so we can combine them.
. . Combine the numerators and "keep" the common denominator.
So we have: .$\displaystyle \frac{c - 2 + 2(c + 2) + 4}{(c+2)(c-2)} \;=\;\frac{c - 2 + 2c + 4 + 4}{(c+2)(c-2)} \;=\;\frac{3c+6}{(c+2)(c-2)}$
Are we done? . . . Well, no. .The fraction can be reduced.
Factor: .$\displaystyle \frac{3(c + 2)}{(c + 2)(c-2)} $
And cancel the $\displaystyle (c+2)\;\; \cdots \;\;\text{Answer: }\;\frac{3}{c-2}$