Dtyjlui

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
$$2x^2 + 5x - 3 = 2x^2 - x + 6x - 3$$

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 $2x^2 + 5x − 3$. Since $a =2$ is prime, we let $d_1 = 2$ and renaming the (unknown) constants, write,

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

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, with the same size spreadsheet, can be used whenever both $a$ and $c$ are prime numbers. In general, you'll have to organize your work and break things down into more cases to find the solution (c.f. the BOX METHOD). However, the approach discussed here, can be extended and methodically applied to handle any of these problems.

Note 1: If you are lazy, just fill in the first row of the spreadsheet for the choices available for $u$, and then use cell formulas:

Note 2: These factorization techniques can come up 'empty handed' - there may be no solutions.

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 polynomial with $2$ and get $$4x^2+10x-6=(2x)^2+5(2x)-6$$ which can be seen as a simple equation in $2x$ with factoring $((2x)+6)((2x)-1)$.

Taking out the added factor $2$ leaves $(x+3)(2x-1).$

For the general quadratic polynomial with integer coefficients $ax^2+bx+c$ the same approach requires the factoring of $a(ax^2+bx+c)=(ax)^2+b(ax)+ac$.

If a factoring $((ax)+m)((ax)+n)$ with integers $m$ and $n$ exists, then general theory guarantees that the extra factor $a$ can be taken out and still leave a factoring with only integer coefficients.

$$2x^2 + 5x − 3 = 0$$

$$ac = 2(-3) = -6$$

$$text{$^-1times 6 =phantom .^-6 $ and $ ^-1+6 = 5$}$$

$-1$ and $6$ are correct. What you did after that is wrong.

Here are two methods that I know of for proceeding from $-1$ and $6$.

Method 1. Replace $5x$ with $-1x+6x$ and factor-by-pairing-off.

begin{array}{c}
2x^2 + 5x − 3 \
2x^2 -1x + 6x - 3 \
(2x^2 -1x) + (6x - 3) \
x(2x-1) + 3(2x-1) \
(x+3)(2x-1)
end{array}

Method 2. Write out $(ax-1)(ax+6)$ and then divide out the greatest common divisors.
begin{array}{c}
(2x-1)(2x+6) &{gcd(2,-1)=1 text{and} gcd(2,6)=2}\
dfrac{(2x-1)}{1} cdot dfrac{(2x+6)}{2} \
(2x-1)(x+3)
end{array}

So we can actually generalize this. Say we have the polynomial
$$p(x)=ax^2+bx+c$$
Fact:
$$p(x)=bigg(ax+frac{b-sqrt{b^2-4ac}}{2}bigg)bigg(ax+frac{b+sqrt{b^2-4ac}}{2}bigg)$$

Proof:

Let's assume the existence of three real numbers $r_1$, $r_2$, and $e_1$ such that
$$ax^2+bx+c=e_1(x-r_1)(x-r_2)$$
If we expand the product on the right hand side and then compare coefficients,
$$ax^2+bx+c=e_1x^2-e_1(r_1+r_2)x+e_1r_1r_2$$
we get a system of equations
$$e_1=a\-e_1(r_1+r_2)=b\e_1r_1r_2=c$$
Evidently, we get $e_1=a$ for free. So we update our system of equations:
$$r_1+r_2=-frac{b}{a}\r_1r_2=frac{c}{a}$$
We can solve each equation for $r_2$:
$$r_2=-frac{b}{a}-r_1\r_2=frac{c}{ar_1}$$
So we can set the two equations equal to each-other:
$$-frac{b}{a}-r_1=frac{c}{ar_1}$$
$$r_1+frac{c}{ar_1}=-frac{b}{a}$$
multiplying both sides by $ar_1$,
$$ar_1^2+br_1=-c$$
Then we add $frac{b^2}{4a}$ to both sides:
$$ar_1^2+br_1+frac{b^2}{4a}=frac{b^2}{4a}-c$$
Then we note that
$$a(r_1+b/2a)^2=ar_1^2+br_1+frac{b^2}{4a}$$
So we plug it in:
$$a(r_1+b/2a)^2=frac{b^2}{4a}-c$$
$$a(r_1+b/2a)^2=frac{b^2-4ac}{4a}$$
$$(r_1+b/2a)^2=frac{b^2-4ac}{4a^2}$$
$$r_1+b/2a=sqrt{frac{b^2-4ac}{4a^2}}$$
$$r_1+b/2a=frac{sqrt{b^2-4ac}}{sqrt{4a^2}}$$
$$r_1+b/2a=frac{sqrt{b^2-4ac}}{2a}$$
$$r_1=frac{-b+sqrt{b^2-4ac}}{2a}$$
And since we know that
$$r_2=-frac{b}{a}-r_1$$
We know that
$$r_2=-frac{b}{a}-bigg(frac{-b+sqrt{b^2-4ac}}{2a}bigg)$$
$$r_2=-frac{2b}{2a}+frac{b-sqrt{b^2-4ac}}{2a}$$
$$r_2=frac{b-2b-sqrt{b^2-4ac}}{2a}$$
$$r_2=frac{-b-sqrt{b^2-4ac}}{2a}$$
And by definition,
$$ax^2+bx+c=abigg(x+frac{b-sqrt{b^2-4ac}}{2a}bigg)bigg(x+frac{b+sqrt{b^2-4ac}}{2a}bigg)$$
$$ax^2+bx+c=bigg(ax+frac{b-sqrt{b^2-4ac}}{2}bigg)bigg(ax+frac{b+sqrt{b^2-4ac}}{2}bigg)$$
And with that our proof is complete :)