Factor 10r^2-9r+2

Question

Factor 10r^2-9r+2

anonymous · Answer

@Hero

anonymous · Answer

I think it would start with (5r+___)(2r+___) but I'm not sure. I thought the second part of each parenthesis has to multiply to equal -9 and add up to 2 but I can't think of anything

anonymous · Answer

$$10r^2-9r+2$$
We need two numbers which add to -9 but multiply to 20

anonymous · Answer

you mean to 2?

anonymous · Answer

where did you get 20 from?

anonymous · Answer

No, in the form of $ax^2+bx+c$ the roots need to multiply to $ac$

anonymous · Answer

Okay, so it would be (5r-5)(2r-4)?

anonymous · Answer

@KeithAfasCalcLover

kc_kennylau · Answer

It is not 20, it is 2

kc_kennylau · Answer

And the answer is (5 r-2) (2 r-1)

anonymous · Answer

Oh i was referring to factoring by decomposition like so:
$$\begin{align}
  10r^2-9r+2&=10r^2-(4+5)r+2 \
  &=10r^2-5r-4r+2 \
  &=5r(2r-1)-2(2r-1) \
  &=(5r-2)(2r-1)
   \end{align}$$

hero · Answer

@Sunshine447 , the way @KeithAfasCalcLover did it is the more preferred method because it is an application of the distributive property. The only thing is @KeithAfasCalcLover should explain it in more detail so that you understand it properly.

anonymous · Answer

Haha of course. In the case of a quadratic function, $f(x)=ax^2+bx+c$, when we analyze the equation: $ax^2+bx+c=0$ we have three cases we need to worry about.

Case 1: $(a=1)$
When $a=1$ we know to use the algorithm of analyzing the $(b)$ and $(c)$ values. For example, with the equation $x^2-5x+4=0$ we usually think: "What two numbers multiply to $+4$ but add to $-5$?" and we reach the conclusion of $\{-4,-1\}$. And so our equation develops to $(x-4)(x-1)=0$

Case 2: $a\in ℝ$
When $a$ becomes any other number, we must develop a different method. This method is known as the factoring by decomposition. With this method, we need to develop our ideas to which two numbers multiply to $a\times c$ but add to $(b)$ So then we follow the method like how I solved your question above. Note how the method degenerates into the same method as case one if our a is, in fact, 1

Our case three is essentially to use the quadratic equation if our b and c values cannot work.