# Factor $6x^2​ −7x−5=0$

I'm trying to factor

$$6x^2​ −7x−5=0$$

but I have no clue about how to do it. I would be able to factor this:

$$x^2-14x+40=0$$ $$a+b=-14$$ $$ab=40$$

But $6x^2​ −7x−5=0$ looks like it's not following the rules because of the coefficient of $x$. Any hints?

• $(2x+1)(3x-5)$ You set it up as $(ax+b)(cx+d)$. wlog $b=\pm1,d=\pm5$ and there are only a few choices for $a,c$. May 8, 2016 at 14:23
• "the rules" should be: find a, b, c, d such that: i) ac = 6 ii) bd = -5 iii) ad + bc = -7. THey are harder but they are still rules. just try all combinations of factors of 6 and 5 and the resulting fractions until you get one that works. ... if you get one that works. It's inadequacy of "the rules" that should encourage you to learn the quadratic equation. May 8, 2016 at 15:55
• use the theorem of Vieta May 8, 2016 at 16:04
• Try to factor $x^2-\frac76x-\frac56$ !
– user65203
Jun 17, 2016 at 17:07
• Similar to math.stackexchange.com/questions/973974/…, just differ by -9.
– user312097
Jun 17, 2016 at 17:10

Consider the product \begin{align*} (4x + 7)(3x - 8) & = 4x(3x - 8) + 7(3x - 8)\\ & = \color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}\\ & = \color{blue}{12}x^2 \color{green}{- 11}x \color{blue}{- 56} \end{align*} Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients in the expression $\color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}$, that is, $$(\color{blue}{12})(\color{blue}{-56}) = (\color{green}{-32})(\color{green}{21}) = -672$$ Now suppose that $ax^2 + bx + c$ is a quadratic polynomial with integer coefficients that has factorization \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\ & = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su}\\ \end{align*} Equating coefficients yields \begin{align*} a & = \color{blue}{rt}\\ b & = \color{green}{ru + st}\\ c & = \color{blue}{su} \end{align*} Note that the product of the quadratic and linear coefficients is equal to the product of the two linear coefficients that have sum $b$, that is, $$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st})$$ Hence, if a quadratic polynomial $ax^2 + bx + c$ factors with respect to the rational numbers, we can split the linear term into two linear terms whose coefficients have product $ac$ and sum $b$.

To split the linear term of $6x^2 - 7x - 5$, we must find two numbers with product $6 \cdot (-5) = -30$ and sum $-7$. They are $-10$ and $3$. Hence, \begin{align*} 6x^2 - 7x - 5 & = 0\\ 6x^2 - 10x + 3x - 5 & = 0 && \text{split the linear term}\\ 2x(3x - 5) + 1(3x - 5) & = 0 && \text{factor by grouping}\\ (2x + 1)(3x - 5) & = 0 && \text{extract the common factor} \end{align*} To solve the equation, use the zero product property $ab = 0 \iff a = 0~\text{or}~b = 0$.

Realize that $$(ax-b)(cx+d)=acx^2-(bc-ad)x-bd$$ Put $a=3$, $c=2$, $b=5$, $d=1$.

$$6x^2-7x-5=3 \times 2 x^2-(2 \times 5 - 3 \times 1)x -5 \times 1=(3 \times x-5)(2 \times x+1)$$

First check whether it factors over the rationals. A quadratic trinomial factors if and only if the discriminant $b^2-4ac$ is a perfect square. $$(-7)^2-4(6)(-5)=169=13^2$$ Next, solve $a+b=-7$ and $ab=(6)(-5)=-30$. So $a=-10$ and $b=3$.

Use these two results to rewrite the $-7x$ as $-10x+3x$.

$$6x^2-10x+3x-5$$

Then factor by grouping:

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

$$ax^2+bx+c=a(x-x_1)(x-x_2),$$ where $$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

For example: $$3x^2-7x+2=0$$ $$x_{1,2}=\frac{7\pm \sqrt{7^2-24}}{2\cdot3}=\frac{7\pm \sqrt{25}}{6}$$ $$x_1=\frac13; x_2=2$$ $$3x^2-7x+2=3(x-\frac13)(x-2)=(3x-1)(x-2)$$

Suppose it has a factorization as a product $$6 x^2 - 7 x - 5 = (a x + b) (c x + d)$$ of linear expressions with rational coefficients. This forces the coefficients to be integers. Multiplying out the right-hand side, we get $$6 x^2 - 7x - 5 = (a c) x^2 + \cdots + b d .$$ So, $a c$ are integers whose product is $6$, and we may as well assume they are both positive (since we can multiply both of the monomials by $-1$ without changing the value of their product). Thus, $a, c$ are either $1, 6$ or $2, 3$, and by relabeling if necessary, the factorization is either $$(6 x + b) (x + d) \qquad \textrm{or} \qquad (3 x + b) (2 x + d) .$$ Now, we also have $bd = -5$, so one of $b$ and $d$ must be $\pm 1$ and the other must be $\mp 5$. We don't know a priori which is which, leaving four candidates for factorizations.

When in doubt, you always have the quadratic equation. But if you really want to do it this way, you have to consider how the dominant term can factor. Here, you have either $6=6\times 1$ or $6=3\times 2$, so $(6x+a)(x+b)$ or $(3x+a)(2x+b)$. Try both. In both cases, you get the Viete-like formulas, just like you wrote down, but with a few prefactors ($a+6b=-7$ and $ab=-5$ in the first case).

We can use the AC method, which states that you have to find two factors that multiply up to $ac$ that also add up to $b$. Note that $ac = -30$. There are two factors that satisfy the condition, -10 and 3 (since $-10 + 3 = 3 - 10 = -7$). Now let $u = 3$ and $v = -10$. We can group according to the following form: $$(ax^2 + ux) + (vx + c)$$ When we do that, we get: $$(6x^2 + 3x) + (-10x - 5)$$ Now, you have to note the common factors of each monomial in the binomial. The GCF for $6x^2 + 3x$ is $3x$, hence we can factor that out: $$3x(2x + 1) + (-10x - 5)$$ Also, the GCF of $-10x - 5$ is -5, so we can factor that out: $$3x(2x + 1) - 5(2x+1)$$ The common term here is $2x + 1$, so we factor that out to get: $$(2x + 1)(3x - 5)$$

You can reduce to the $x^2+Ax+B$ case with a trick: multiply everything by $6$, because you want the leading coefficient to be an even square. We'll remove the factor $6$ later. You get $$36x^2-42x-30$$ Now you can set $y=6x$, so the polynomial can be written $$y^2-7y-30$$ and you need to find two numbers whose sum is $7$ and whose product is $-30$, which are $10$ and $-3$. So you have $$y^2-7y-30=(y-10)(y+3)$$ and your original polynomial is (also restoring the factor $1/6$) $$\frac{1}{6}(6x-10)(6x+3)=\frac{2\cdot 3}{6}(3x-5)(2x+1)=(3x-5)(2x+1)$$

Here is a proof from Sierpinski, Waclaw: Elementary theory of numbers

Let us assume that $a_m,\ldots,a_0$ are integers and $x=\frac{k}{s}$ is a solution of $$a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0=0 \tag{1}$$ with $$(k,s)=1 \tag{2}$$ then

$$a_mk_m+a_{m-1}k^{m-1}s+\cdots+a_1ks^{m-1}+a_0s^{m}=0 \tag{3}$$ and therefore
$$a_mk^m=-(a_{m-1}k^{m-1}+\cdots+a_1ks^{m-2}a_0s_{m-1})s \tag{4}$$ and $$a_0s^{m}=-(a_mk_{m-1}+a_{m-1}k^{m-2}s+\cdots+a_1s^{m-1})k \tag{5}$$

From $(3)$ follows $s|a_mk^m$, and because of $(2)$, we have $s|a_m$. From $(4)$ follows $k|a_0s^{m}$ and because of $(2)$, we have $k|a_0$.

So the rational solutions $x=\frac{k}{s}$ of $(1)$ can be found by checking all possible $k|a_0$ and $s|a_m$.

Here's an alternative (but you should just learn the quadratic equation)

$6x^2 - 7x - 5= 6(x + a)(x + b)$

So $ab = -5/6$ and $a + b = -7/6$.

$a = \pm 1, \pm 5, \pm 1/2, \pm 5/2, \pm 1/3, \pm 5/3, \pm 1/6, \pm 5/6$

$b = \mp 5/6, \mp 1/6, \mp 5/3, \mp 1/3, \mp 5/2, \mp 1/2, \mp 5, \mp 1$.

So $a + b = \pm 2/3, \pm 29/6, \mp 7/6, \pm 13/6, etc....$

So $a = 1/2$ and $b= -5/3$ works.

$6(x + 1/2)(x - 5/3) = (2x + 1)(3x - 5) = 6x^2 - 7x -5$.