Factorization of Quadratic Polynomials

Factorization of Quadratic Polynomials

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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:

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

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