How to see $2n^2 + 7n +6 = (n+2)(2n+3)$?

Question

In a proof I was reading of the sum of the first $n$ squares formula, the fact is used that $2n^2 + 7n +6 = (n+2)(2n+3)$. I was struggling with the exact logic used in deducing this relation- I see we could use the quadratic formula to find the roots $-2$ and $-3 \over 2$; but wasn't seeing the exact logic used to get $(n+2)(2n+3)$. I would appreciate an overview of the exact logic used here, such as which theorems we are availing ourselves of, etc.

Answer 1

Sometimes there is a shortcut:
Given integers $a, b, c$, sometimes it is easy to find integers $b_1, b_2$ s.t. $b=b_1+b_2, ac=b_1\times b_2$. Then the factorisation is easily done by splitting the middle term like so: $ax^2+bx+c=ax^2+b_1x+b_2x+b_1b_2/a = ax(x+b_1/a)+b_2(x+b_1/a)=(ax+b_2)(x+b_1/a)$.

In particular if you can select $b_1$ s.t. $a\mid b_1$ then the coefficients remain integer.

In the e.g. in question, it is quickly noted that $b=7=4+3, ac=12 = 4\times 3$. Hence $2n^2+7n+6 = 2n^2+4n+3n+6=2n(n+2)+3(n+2)=(2n+3)(n+2)$

Answer 2

I'm not sure where this is used in the proof of sum of first n squares formula but factorizing an expression in x is a very common and useful thing to for several purposes such as-

• Getting to know for what values of x the expression is +ve, -ve, or zero.
• Simplifying the calculations for evaluating the value of a function for larger values of x (this is not always the case as sometimes it might be easier to put the value in the expression as it is) So here the utility varies as per parameters such as the value of x and the expression at hand.
• Factorization is also immensely useful in solving Diophantine equations in Number Theory where we can use property of integers or naturals to get the values of the variables.

These are all I could recall at the moment but there are several other uses you would come across in various domains of Mathematics and the Sciences

So, in conclusion, Factorization is basically providing an equivalent of the expression in a different format so that you can analyze it in different ways. Keep in mind that the original 'unfactorized' form is also very useful while finding integrals and derivatives of the expression.

P.S. - This is my first answer on this website so excuse me if I made any mistakes while answering the question. Feel free to share your suggestions on how to write better answers

Answer 3

If you want a systematic procedure try completing the square. This is a useful procedure in its own right. Start by factoring out the 2

$2n^2+7n+6=2(n^2+\frac{7}{2}n+3)$

Now using $(a+b)^2=a^2+2ab+b^2$ so that $a^2+2ab=(a+b)^2-b^2$ you can write

$n^2+\frac{7}{2}n+3=(n+\frac{7}{4})^2-(\frac{7}{4})^2+3=(n+\frac{7}{4})^2-\frac{1}{16}$

This is the completed square form of $n^2+\frac{7}{2}n+3$. Now we can apply the difference of two squares formula $a^2-b^2=(a-b)(a+b)$ to get

$(n+\frac{7}{4})^2-\frac{1}{16}=(n+\frac{7}{4}-\frac{1}{4})(n+\frac{7}{4}+\frac{1}{4})=(n+\frac{3}{2})(n+2)$

So that

$2n^2+7n+6=2(n+\frac{3}{2})(n+2)$