# Factorization of Quadratic Polynomials

How can we factorize a *quadratic polynomial into two linear factors* (if it is possible)? We have already seen one way: use the Factor Theorem to figure out the factors of the polynomial. This is more of a hit-and-trial method as we have to input different values into the polynomial and see if we are able to generate 0. For example, consider the polynomial

\(p\left( x \right):2{x^2} - 6x + 4\)

We note that \(p\left( 1 \right) = p\left( 2 \right) = 0,\)which means that\(\left( {x - 1} \right),\,\,\left( {x - 2} \right)\) are factors of \(p\left( x \right)\). Thus, we can write

\[p\left( x \right) = \lambda \left( {x - 1} \right)\left( {x - 2} \right)\]

If we compare the coefficients of \(p\left( x \right)\) and the right hand side above, we can easily conclude that \(\lambda = 2,\) and so

\[p\left( x \right) = 2\left( {x - 1} \right)\left( {x - 2} \right)\]

Now, consider the polynomial

\[q\left( x \right):4{x^2} + 24x + 35\]

In this case, it is much harder to figure out those (two) values of *x* which, when we substitute into this polynomial, are able to generate 0. We therefore do something else; we *split the middle term*, as follows:

\[\begin{array}{l}p\left( x \right) = 4{x^2} + 24x + 35\\\;\;\;\;\;\;\;\; = 4{x^2} + 14x + 10x + 35\end{array}\]

Now, we take out common factors from the first two and the last two terms:

\[\begin{array}{l}p\left(x\right)=2x\left(2x+7\right)+5\left(2x+7\right)\\\;\;\;\;\;\;\;\;\,=\left(2x+5\right)\left(2x+7\right)\end{array}\]

Thus, we have succeeded in finding out the factors of \(p\left( x \right)\). But how did we figure out how exactly to split the middle term? Let us assume that *a*, *b*, *c* and *d* are integers such that

\[\begin{array}{l}4{x^2} + 24x + 35 = \left( {ax + b} \right)\left( {cx + d} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, = ac{x^2} + adx + bcx + bd\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, = ac{x^2} + \left( {ad + bc} \right)x + bd\end{array}\]

Our problem is to figure out these four integers (we already know the answer, but we are trying to develop a methodology to be able to figure out the answer in any case). Comparing the coefficients above, we see that

\[\begin{array}{l}\left. \begin{array}{l}ac = 4\\bd = 35\end{array} \right\} \Rightarrow abcd = 140\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ad + bc = 24\end{array}\]

Thus, we have to figure out two integers, namely \(ad,\,\,bc\) whose product is

\(\left( {ad} \right)\left( {bc} \right) = 140\)

and sum is

\(ad + bc = 24\)

By observation, we can say that

\(ad = 14,\,\,bc = 10\)

Thus, we split the middle term as:

\(24 = 14 + 10\)

This method of splitting the middle term works well only in cases where the coefficients are small integers and it’s easy to guess by observation how to split the term.

**Example 1:** Factorize the following polynomial by splitting the middle term:

\[p\left( x \right):{x^2} + 10x + 24\]

**Solution:** Since the coefficient of \({x^2}\) is 1, we can assume that in both the linear factors, *x* will have a coefficient of 1. Thus, we have to find integers *a* and *b* such that

\[\begin{array}{l}{x^2} + 10x + 24 = \left( {x + a} \right)\left( {x + b} \right)\\ \;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\!\!\!\; = {x^2} + \left( {a + b} \right)x + ab\end{array}\]

Comparing the coefficients on both sides, we have

\(\begin{array}{l}a + b = 10\\\,\,\,\,\,ab = 24\end{array}\)

We thus have to figure out two integers whose sum is 10 and product is 24. A little thinking shows that we can take

\(a = 6,\,\,b = 4\)

Accordingly, we split the middle term of \(p\left( x \right)\) and obtain the two factors:

\[\begin{array}{l}\,{x^2} + 10x + 2 = {x^2} + 6x + 4x + 24\\ \;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;= x\left( {x + 6} \right) + 4\left( {x + 6} \right)\\ \;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;= \left( {x + 6} \right)\left( {x + 4} \right)\end{array}\]

**Example 2:** Factorize the following polynomial by splitting the middle term:

\[p\left( x \right):8{x^2} + 30x + 7\]

**Solution:** We have to figure out two integers whose sum is 30 and whose product is

\(8 \times 7 = 56\)

These two integers are 28 and 2; we can now split the middle term accordingly and obtain the two factors:

\[\begin{array}{l}8{x^2} + 30x + 7 = 8{x^2} + 2x + 28x + 7\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;= 2x\left( {4x + 1} \right) + 7\left( {4x + 1} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\; \;\;= \left( {2x + 7} \right)\left( {4x + 1} \right) & \end{array}\]

**Example 3:** Factorize the following polynomial by splitting the middle term:

\[p\left( x \right):30{x^2} + 23x + 4\]

**Solution:** We have to figure out two integers whose sum is 23 and whose product is \(30 \times 4 = 120\). These two integers are 15 and 8; we can now split the middle term accordingly and obtain the two factors:

\[\begin{array}{l}30{x^2} + 23x + 4 = 30{x^2} + 15x + 8x + 4\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;= 15x\left( {2x + 1} \right) + 4\left( {2x + 1} \right)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;= \left( {15x + 4} \right)\left( {2x + 1} \right)\end{array}\]