Factoring $2x^2 - 7x-4$

Question

This is an example from the Stewart's Review of Algebra - https://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/ess-reviewofalgebra.pdf

It is on page $3$, example $5$. This is what it suggests on the book

Even though the coefficient of $x^2$ is not $1$, we can still look for factors of the form $2x+r$ and $x+s$, where $rs=-4$. Experimentation reveals that: $2x^2-7x-4=(2x+1)(x-4)$

Now, I have no idea what they mean by "experimentation" there. I had $2$ ways to find the solution of this problem: Use Vieta's Theorem or divide the equation by $2$ so the coefficient of a is $1$, factor it using $(x+r)(x+s)$ and then add the coefficient $2$ back. I still have a feeling that none of the methods I tried were the one suggested by the author. Can anyone explain what the author meant by "experimentation" and how he came to that solution?

Answer 1

If$$2x^2-7x-4=(2x+r)(x+s)=2x^2+(r+2s)x+rs,$$then $rs=-4$ and $r+2s=-7$. I suspect the what the author is suggesting here is to take all pairs $(r,s)$ of integers such that $rs=-4$ (there are only $6$ such pairs), and to see whether or net we have $r+2s=-7$ for one such pair. And the answer is affirmative: the pair $(1,-4)$ will work.

Answer 2

I guess you are beginning - where you have reason to suspect the factors are likely to be "nice" (here meaning that the roots are rational) you can assume $(2x+r)(x+s)$ with $r$ and $s$ integers (not just rational numbers). If you are wrong in your assumption, you have extra work.

Whether the answer is considered correct depends on the precise question. If you are asked to find a factorisation then any method (including trial and error will do).

If the equation $px^2+qx+r=0$ with $p, q, r$ being integers and $p$ a prime has rational roots, then one of the roots is an integer factor of $r$ (may be positive or negative). This is a consequence of the fact that $p$ is a prime. If $\alpha$ is a root, then $(x-\alpha)$ is a factor. So, since $2$ is a prime, that is easy.

Answer 3

If the given quadratic is factorable (over integers), to get the $x^2$-term to come out right, you would need $(2x...)(x...)$. Because the given quadratic has no common integer factor (greater than $1$) of the three coefficients, we must have that neither factor in the factorization has a common integer factor. So there will have to be $\pm 1$ with the $2x$ factor, and correspondingly $\mp 4$ with the $x$ factor. So there are only two possibilities to check.

(This goes faster with practice...a lot faster than the explanation!)

Answer 4

Here is a method for factorising quadratics that reduces the amount of trial and error needed. To factorise any quadratic $ax^2+bx+c$, you need to find two numbers that have a sum of $b$ and a product of $ac$. In your example of $2x^2-7x-4$, we need to find two numbers that sum to make $-7$ and multiply to make $-8$. Inspection tells us that those numbers are $-8$ and $+1$. Then, you should split up the middle term of the quadratic like so: \begin{align} 2x^2\color{red}{-7}x-4 &= 2x^2\color{red}{-8}x+\color{red}{1}x-4 \\[5pt] &= 2x(x-4)+(x-4) \\[5pt] &= (2x+1)(x-4) \, . \end{align} See this article for more information. Ask me if you have any questions.