Tukukkk

A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?

When following the factors $-1$ and $6$. I have

$(2x^2-1x)(6x-3)$

$x(2x-1)+3(2x-1)$

Is this correct, if not; what is the best way to solve a leading coefficient when factoring?

Yes your solution is correct.

Further, note that if you know how to solve a quadratic equation, it's much more easier to factorise in those cases. Suppose the equation is $ax^2+bx+c=0$, and you can find it's roots, $x_1,x_2$. Then, you can simply write $$ax^2+bx+c=a(x-x_1)(x-x_2)=0$$

We need to find two numbers with product $2(-3) = -6$ and sum $5$. You are correct that the numbers are $-1$ and $6$. Splitting the linear term yields
begin{align*}
2x^2 + 5x - 3 & = 2x^2 - x + 6x - 3 && text{split the linear term}\
& = x(2x - 1) + 3(2x - 1) && text{factor by grouping}\
& = (x + 3)(2x - 1) && text{extract the common factor}
end{align*}
which you can verify by multiplying the factors.

You should not write $(2x^2 - 1x)(6x - 3)$ since
begin{align*}
(2x^2 - 1x)(6x + 3) & = 2x^2(6x + 3) - 1x(6x + 3)\
& = 12x^3 + 6x^2 - 6x^2 -3x\
& = 12x^3 - 3x\
& neq 2x^2 + 5x - 3
end{align*}
Instead, you can write $(2x^2 - 1x) + (6x - 3)$ or $2x^2 - 1x + 6x - 3$.

Also, you should be including equals signs since you are asserting that
$$2x^2 + 5x - 3 = 2x^2 - x + 6x - 3$$

We have that $(2x+a)(x+b)=2x^2+5x-3$, where $x=-frac a2$ and $x=-b$ are the solutions you require.

Then $$(2x^2+(a+2b)x+ab=2x^2+5x-3$$
leads to $$a+2b=5; ab=-3$$
We set $b=frac12(5-a)tofrac12a(5-a)=-3 to a^2-5a-6=0$
$$to a=-1, 6$$
$$to b= 3, -frac 12$$

So $$2x^2+5x-3=(2x-1)(x+3)=(2x+6)(x-frac12)$$

One could start with multiplying the equation with $2$ and get $$4x^2+10x-6=(2x)^2+5(2x)-6=0$$ which can be seen as a simple equation in $2x$.

It is worth reviewing the theory behind the OP's technique.

Assume always that $a$ is a positive integer, $a ge 1$.

Suppose

$tag 1 ax^2+bx+c= (d_1x+e_1)(d_2x+e_2)$

where the constants $b, c, d_1. e_1, d_2 text{ and } e_2$ are all integers.

If $text{(1)}$ holds true then we can write

$tag 2 ax^2+bx+c= (d_1'x+e_1')(d_2'x+e_2')$

with integer constants $d_1', e_1', d_2' text{ and } e_2'$ such that $d_1' text{ and } d_2'$ are both positive.

Hint: If necessary, apply $(-1)(-1) = 1$ to the rhs of $text{(1)}$

Now we are looking for a '$text{(1)}$ factorization' of $2x^2 + 5x − 3$,

$tag 3 2x^2 + 5x − 3= (2x+u)(x+v)=2x^2 + (2v+u)x + uv$

Noticing that $uv = -3$ restricts things considerably, we create a google spreadsheet,

and find the answer:

$tag 4 u = -1 text{ and } v = 3$

so

$tag 5 2x^2 + 5x − 3= (2x-1)(x+3)$

This technique can be used whenever both $a$ and $c$ are prime numbers.

Note: If you are lazy, just fill in the first row of the spreadsheet for the choices available for $u$, and then use cell formulas: