Factorize $6x^2 -5x -14 = 0$

Question

I'm throwing a bit of a blank on the best way to factor this :

$$6x^2 -5x -14 = 0$$

I know that I could multiply $6$ by $14$ and then find a pair of factors that add to $-5$ (b), but this feels a bit convoluted and I'm sure that I've gone about this a bit of an easier way before, I just can't remember how...

Does anyone know the simplest way?

cheers!

Answer 1

The simplest way is to factor by splitting and grouping. To split the linear term of $6x^2 - 5x - 14$, we must find two numbers with product $6 \cdot -14 = -84$ and sum $-5$. They are $-12$ and $7$. Hence,

\begin{align*} 6x^2 - 5x - 14 & = 0\\ 6x^2 - 12x + 7x - 14 & = 0 && \text{split the linear term}\\ 6x(x - 2) + 7(x - 2) & = 0 && \text{factor by grouping}\\ (6x + 7)(x - 2) & = 0 && \text{extract the common factor} \end{align*}

If a product is equal to zero, then one of the factors must equal zero. Hence,

\begin{align*} 6x + 7 & = 0 & x - 2 & = 0\\ 6x & = -7 & x & = 2\\ x & = -\frac{7}{6} \end{align*}

Note that if you want to check your factorization is correct, you form the product

\begin{align*} (6x + 7)(x - 2) & = 6x(x - 2) + 7(x - 2)\\ & = 6x^2 - 12x + 7x - 14\\ & = 6x^2 - 5x - 14 \end{align*} so factoring by splitting and grouping is simply performing the steps of multiplying two linear factors in reverse order. The reason it works is that if a quadratic polynomial splits into linear factors, then the product of the quadratic coefficient and constant term is equal to the product of the coefficients of the two linear terms into which it splits. In this case, $6(-14) = (-12)(7)$. In general, if

$$ax^2 + bx + c = (mx + n)(px + q) = mpx^2 + (mq + np)x + nq$$

then $ac = (mp)(nq) = (mq)(np)$, where $mq + np = b$.

This is certainly more efficient than guessing the linear factors. You could find the factors directly by completing the square (as Michael Hardy demonstrated), but this involves more work. You could also find the factors by solving the equation using the Quadratic Formula, then working backwards to determine the factors (again, as Michael Hardy demonstrated). However, if your goal is to solve the equation, you do not need to factor the quadratic if you have applied the Quadratic Formula to find the roots.

I will step off my soapbox now.

Answer 2

You probably got the quickest way. Another way is to multiply by $4\times 6 =24$ - where you would always have the $4$ and the $6$ would be the coefficient of $x^2$. This gives $$4\cdot6^2\cdot x^2-4\cdot 6 \cdot 5\cdot x-4\cdot 6\cdot 14=(12x-5)^2-25-336=(12x-5)^2-19^2=0$$ which keeps everything integral, and if there is a rational factorisation will come out nicely as the difference of two squares. If not, you get an irrational square root and it is wholly equivalent to the usual quadratic formula.

Here the difference of two squares factorises as $(12x-24)(12x+14)$ corresponding to $(x-2)(6x+7)$ for the original form.

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In general starting with $ax^2+bx+c=0$ would go via $4a^2x^2+4abx+4ac=(2ax+b)^2-b^2+4ac=0$ and will come out nicely if $b^2-4ac$ is a square. The main feature of this method is that it completes the square without introducing fractions.

Answer 3

\begin{align} & 6x^2 - 5x - 14 \\[8pt] = {} & 6\left( x^2 - \frac 5 6 x \right) - 14 \\[8pt] = {} & 6\left( x^2 - \frac 5 6 x + \frac{25}{144} \right) - 14 - 6 \left(\frac{25}{144}\right) \quad\text{(completing the square)} \\[8pt] = {} & 6 \left(\left( x - \frac 5 {12} \right)^2 - \frac{361}{144}\right) \\[8pt] = {} & 6 \left(\left( x - \frac 5 {12} \right)^2 - \left(\frac{19}{12}\right)^2 \right) \\[8pt] = {} & 6\left(x - \frac5{12} - \frac{19}{12}\right)\left(x - \frac5{12} + \frac{19}{12}\right) \quad\text{(factoring a difference of two squares)} \\[8pt] = {} & 6(x-2)\left(x + \frac 7 6\right) \\[8pt] = {} & (x-2)(6x+7). \end{align}

If one says $$ x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{5\pm\sqrt{5^2-4\cdot6\cdot(-14)}}{2\cdot6} = \frac{5\pm19}{12} = \left(\frac{24}{12}\text{ or }\frac{-14}{12}\right) = \left(2\text{ or }\frac{-7}6\right) $$ then one knows that $x-2$ and $x-\left(\dfrac{-7}6\right)$ must be factors, and one must have $6$ as the leading coefficient, so one gets the same result.

The usual way to prove that $$ x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a} $$ if $ax^2+bx+c=0$ is by completing the square.