Factorize $2x^2 - 5x + 3$

Question

How do I factorize that because I can't simplify my quadratic fraction without factorizing that part

Here's my full question if your curious:

Simplify fully : $$\frac{x^2 + 3x - 4}{2x^2 - 5x + 3}$$

Answer 1

Assume: $$\frac {x^2+3x-4}{2x^2-5x+3}$$

We can write this like:

$$\frac {(x+4)(x-1)}{(x-1)(2x-3)}$$

Answer 2

You can also use the quadratic formula. If your quadratic polynomial is $ax^2+bx+c$, then your two solutions are:

$$S_1 = \frac{-b + \sqrt{b^2-4ac}}{2a}$$

and

$$S_2 = \frac{-b - \sqrt{b^2-4ac}}{2a}.$$

Once you have these, then the polynomial factors to $(x-S_1)(x-S_2).$

With your numerator $x^2+3x-4$, $S_1 = \left[-3+\sqrt{9 - 4(1)(-4)}\right]/[2(1)] = (-3+5)/2 = 1$. This gives you the factor $(x-1)$.

The rest are similar.

Answer 3

If you can factorize $2x^2-5x+3$, you can write it as $c(x-a)(x-b)$. This means that $$2x^2-5x+3=cx^2-c(a+b)x+abc$$

Comparing we see that $c=2$, $a=1$ and $b=\frac{3}{2}$. So $2(x-1)(x-\frac{3}{2})=(x-1)(2x-3)$.

Answer 4

Hint: The presence of the $2$ and the $3$ limit your choices to just a few possibilities, because you can only factor these as $2\cdot1$ and $3\cdot 1$. So it has to be one of $$(2x\pm 1)(x\pm 3)$$ $$(2x\pm 3)(x\pm 1)$$ Note that the signs must match to give $+3$ instead of $-3$.

Also, since the coefficient of the middle term is negative ($-5x$), the signs can't all be positive.

Answer 5

1) Guessing:

It's harder to write it out than to actually do in our heads: "The factors of $2$ are $1$ and $2$ and the factors of $3$ are $1$ and $3$; which factors when combined add to plus or minus $5$? Well, obviously $2+3 = 5$ so $-2-3=-5$ and the factors are $(2x-3 )(1x-1)$".

But if we had to explain or teach it:

$(ax + b)(cx + d) = acx^2 + (bc + ad)x + bd$.

So we need $ac = 2$. So $(a,c) = (1,2)$ or $(2,1)$ or $(-1,-2)$ or $(-2,-1)$.

We might as well fix $a = 1; c = 2$. (We can just switch labels [$(ax+b)(cx+d) = (cx+d)(ax + b)$ so we can switch the $a$ with the $c$] or multiply both terms by $-1$ [$(ax + b)(cx+d) = (-ax-b)(-cx -d)$ so we can switch $a$ with $-a$] so we can "declare $a = 1; c=2$.)

We need $bd = 3$. So $(b,d) = (1,3),(3,1),(-1,-3)$ or $(-3,-1)$

And we need $(bc + ad) = 2b + d = -5$. If $(b,d) = (1,3),(3,1),(-1,-3)$ or $(-3,-1)$ then $2b + d = 5, 7, -5, -7$. So $b = -1$ and $d=-3$ and

$2x^2 -5x + 3 = (x-1)(2x -3)$.

2) Completing the square:

If we can find the roots of $2x^2 -5x + 3 = 0$ are $x= r_1$ and $x = r_2$ then $2x^2 - 5x + 3= 2(x-r_1)(x- r_2)$. (Because if $2x^2 - 5x + 3 = 2(x+ b)(x+d)$ then $2r_i^2 - 5r_i + 3 = 0$ means $(r_i + b)=0$ or $(r_i+d) = 0$ which means the roots are $-b$ and $-c$.)

So $2x^2 - 5x + 3 = 0$

$x^2 -\frac 52 x = -\frac 32$

$x^2 - 2\frac 54x + \frac {25}{16} = \frac {25}{16} - \frac 32 = \frac {25 - 3*8}{16} = \frac 1{16}$.

$(x-\frac 54)^2 = \frac 1{16}$

$x - \frac 54 = \pm \frac 14$

$x = \frac 54 \pm \frac 14 = \frac 64$ or $\frac 44= \frac 32$ or $1$

So $2x^2 - 5x +3 = 2(x - \frac 32)(x -1) = (2x -3)(x -1)$

3)Quadratic formula:

$2x^2 -5x + 3 = 0 \implies x = \frac {5 \pm \sqrt{5^2 - 4*3*2}}{2*2} = \frac {5\pm 1}{4} = \frac 23$ or $1$.

So $2x^2 - 5x +3 =2(x-\frac 32) (x -1) = (2x-3)(x-1)$.