1. ## Factoring

is it possible to factor this??

$\displaystyle 20s^2-49st+30t^2$

the closest i can get is -30 times -20 but the sum of that equal -50

2. Rewrite as:

How about -24 and -25. -24-25=-49, (-24)(-25)=600

$\displaystyle 20s^{2}-24st-25st+30t^{2}$

$\displaystyle 4s(5s-6t)-5t(5s-6t)$

$\displaystyle (4s-5t)(5s-6t)$

3. Hello, t-dot!

Is it possible to factor this: .$\displaystyle 20s^2-49st+30t^2$
If you're desperate, you can use the Quadratic Formula.

Consider the quadratic: .$\displaystyle 20x^2 - 49x + 30 \:=\:0$

Quadratic Formula: .$\displaystyle x \;=\;\frac{-(-49) \pm \sqrt{(-49)^2 - 4(20)(30)}}{2(20)} \;=\;\frac{49 \pm \sqrt{1}}{40}$

And we have: .$\displaystyle x \;=\;\frac{49\pm1}{40} \;=\;\begin{array}{ccccc}\dfrac{49+1}{40} & = & \dfrac{50}{40} & = & \dfrac{5}{4} \\ \dfrac{49-1}{50} & = & \dfrac{48}{40} & = & \dfrac{6}{5}\end{array}$

Now we can run the problem backwards:

. . $\displaystyle x \:=\:\frac{5}{4}\quad\Rightarrow\quad 4x \:=\:5\quad\Rightarrow\quad 4x-5 \:=\:0$

. . $\displaystyle x \:=\:\frac{6}{5}\quad\Rightarrow\quad 5x \:=\:6\quad\Rightarrow\quad 5x-6\:=\:0$

So the equation was: .$\displaystyle (4x-5)(5x-6) \;=\;0$

. . and the quadratic is: .$\displaystyle 20x^2 - 49x + 30 \;=\;0$

Therefore: .$\displaystyle 20s^2 - 49st + 30t^2 \;=\;(4s-5t)(5s-6t)$

4. Originally Posted by Soroban
Hello, t-dot!

If you're desperate, you can use the Quadratic Formula.

Consider the quadratic: .$\displaystyle 20x^2 - 49x + 30 \:=\:0$

Quadratic Formula: .$\displaystyle x \;=\;\frac{-(-49) \pm \sqrt{(-49)^2 - 4(20)(30)}}{2(20)} \;=\;\frac{49 \pm \sqrt{1}}{40}$

And we have: .$\displaystyle x \;=\;\frac{49\pm1}{40} \;=\;\begin{array}{ccccc}\dfrac{49+1}{40} & = & \dfrac{50}{40} & = & \dfrac{5}{4} \\ \dfrac{49-1}{50} & = & \dfrac{48}{40} & = & \dfrac{6}{5}\end{array}$

Now we can run the problem backwards:

. . $\displaystyle x \:=\:\frac{5}{4}\quad\Rightarrow\quad 4x \:=\:5\quad\Rightarrow\quad 4x-5 \:=\:0$

. . $\displaystyle x \:=\:\frac{6}{5}\quad\Rightarrow\quad 5x \:=\:6\quad\Rightarrow\quad 5x-6\:=\:0$

So the equation was: .$\displaystyle (4x-5)(5x-6) \;=\;0$

. . and the quadratic is: .$\displaystyle 20x^2 - 49x + 30 \;=\;0$

Therefore: .$\displaystyle 20s^2 - 49st + 30t^2 \;=\;(4s-5t)(5s-6t)$

thanks but i'm not quite sure i understand this way of solving it.