Question

factor 25x^2 −15x+2

Answer 1

Ok, slightly harder this one.

When approaching quadratics if you can do it in your head that's great! But if you can't you can always use the quadratic formula (google it) which will give you the answers.\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]for eqation \(ax^2+bx+c = 0\)

For this question however we have \[25x^2 −15x+2\]So our answer will be of the form\[(5x +m)(5x+n)\]We know that \(m \times n = 2\)So they can only really be 2 and 1. So let's experiment\[(5x +1)(5x+2)\]but when we multiply that out we get \[25x^2 + 15x+2\]So clearly we have the signs wrong. This implies the answer \[(5x -1)(5x-2)\]Mega long winded way of explaining it, but hope that helps! Let me know if you have any problems.

Answer 2

Why was that harder? \(25=5\times 5\), and the 2 makes one of them into 10, so -15 is possible. Well, I have been doing a lot of factoring lately... so they have started popping out at me.

That is one thing about math. The more you work with the numbers, the more it makes sense.

Answer 3

I too gained knowledge from the above discussion Thanks for it.

Answer 4

There's a somewhat simpler way of doing these which requires (a bit) less trial and error. If you can learn it or a method similar, it'll make factoring much quicker... trial and error freaking sucks.
\[\Large ax^2 +bx+c\]\[\large 25x^2 −15x+2\]
First multiply the a (which is 25) by c (which is 2), 25x2 = 50.

Now find factors of that which add up to b (which is -15)... these are -10 and -5, since (-10)(-5) = 50.

Now write them in like so, just put ax in both sets of parentheses for now: \[\large (25x-10)(25x-5)\]
Now, remember we multiplied by 25 back in the first step? So we have to divide that whole thing by 25, but in this case we'll have to divide one by 5 and the other by 5, since neither set of brackets is divisible by 25:
\[\Large \frac{(25x-10) }{ 5 }\frac{ (25x-5 )}{ 5 }\]giving us \[\Large (5x-2)(5x-1)\]

It's a bit easier to explain in person, but I find it easier to apply than most other methods for factoring ax^2+bx+c when a is greater than 1.

Answer 5

@agent0smith I like that one!

Answer 6

Thanks :) All the students I've shown it to, even ones who aren't very good with math, pick it up pretty quickly.

It's easy to remember - you just multiply the first number by the last number and go from there.

Eg:

\[\Large 3x^2 + 16x - 12\]
3(-12) = -36

18 and -2 add up to 16... and18(-2) = -36, so:

\[\Large (3x-2)(3x+18)\] since we multiplied by 3 at the start, now divide it by 3 (divide the right brackets by 3)\[\Large \frac{ (3x-2)(3x+18) }{ 3 } = (3x-2)(x+6)\]