I am having a hard time on figuring out how to factor this equation.
It says to solve this quadratic equation for "x" by factoring and to leave the answer as exact values.
This is the equation needed to be factor;
2x(squared) -7x + 5 = 0.
Could anyone help me?
HallsofIvy said:
Could anyone help me? You know, I presume, that (ax+ b)(cx+ d)= acx^2+(ad+ bc)x+ bd.
Further, you want these numbers, a, b, c, and d to be integers. So you want ac= 2.
That means you must have a= 2, c= 1 or a= 1, c= 2, or a= -2, c= -1, or a= -1, c= 2,
four possible combinations. You want bd= 5 so you must have b= 5, d= 1, or b= 1, d= 5, or b= -5, d= -1, or b= -1, d= -5.
There are 4(4)= > > > 16 < < < possible combinations of those. Find the one that makes ad+ bc= -7.
.
Here is another of several other alternative methods:
You can use a modified trial and error approach to get it down to two combinations to choose from for the answer.
\(\displaystyle 2x^2 - 7x + 5 = 0\)
Note: Keep in mind that we are factoring over integer coefficients.
The second sign of the trinomial, reading left to right, is a plus sign. That indicates the signs inside the binomial factors
will be the same. In conjunction with that, the first sign of the trinomial, the subtraction sign, indicates that both signs
in the binomial factors will be subtraction signs.
If you concentrate on breaking apart the \(\displaystyle \ 2x^2 \ \) term first into monomial factors (each with a postive
integer coefficient), there is only one way, not including their order. In some order, they split into \(\displaystyle \ 2x \ \ and \ \ x.\)
So far we have this set-up:
\(\displaystyle (2x \ - \ \ \ \ )(x \ - \ \ \ \ ) \ = 0\)
Given that we have decided to anchor down the factoring of the \(\displaystyle \ 2x^2 \ \) term first, then the factoring of the
constant term, 5, in the trinomial, just needs to be decided. As 5 can only be broken down into 1 and 5, when limiting
to positive factors, you need to decide where the factors of 5 get placed. The presence of the "2x" and the "x" in the
binomial factors make the factors asymmetrical in that aspect, so that is why there are two choices for their placement.
