Today we will see how to factorise a second order polynomial $P(x)$ of the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are known coefficients. These form of polynomials are also known as $Quadratic$ polynomial.
One of the easy way or factoring this polynomial is to split up the middle term ($bx$) into two terms such that the product of the middle term is equal to the two terms such that the product of the two splits equal to the product of the first term ($ax^2$) and the third term ($c$).
That is we write:
$ax^2 + bx + c = ax^2 + b_1x + b_2x + c$, such that:
$ax^2 \times c$ = $b_1x \times b_2x$.
We need make sure that the above relations hold considering the signs of the coefficients $a, b$ and $c$.
We will take a few examples to understand this method.
Let us factorise the polynomial $4x^2 - 7x - 15$
We can see that the product of the first and last term, is $4x^2 \times (-15) = -60x^2$, so we need to split the term $-7x$ such that their product is $-60x^2$
We know that $12 \times 5 = 60$ and $-12 + 5 = -7$
So we can split the middle term as $-7x = -12x + 5x$
Thus, we get:
$4x^2 -7x -15$
$= 4x^2 - 12x + 5x - 15$, taking $4x$ common in the first two terms and $5$ common in the third and the fourth term, we get,
$= 4x(x-3) +5(x-3)$. Now we have two terms in our expression, and we can see that $(x-3)$ is common in both the terms. So, we take $(x-3)$ common, and we get:
$= (x-3)(4x+5)$
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Now our expression has only one term with two factors, and we have factored our polynomial as required.
Let us look at one more property of a polynomial factor.
Let is say that $x + a$ is a factor of polynomial $P(x)$
then we can write $P(x) = (x+a) \times G(x)$
If we put $x = -a$ then $x + a = 0$,
and we get $P(x) = (x + a) \times G(x) = 0 \times G(x) = 0$
We can use this property to find out if a given term $(x + a)$ is a factor of a given polynomial $P(x)$. All we need to do is evaluate $P(x)$ for $x = -a$ and check if the result is $0$.
We will understand this better by using the previous example with $P(x) = 4x^2 -7x - 15$
Is $x - 3$ a factor of $4x^2 - 7x -15$?
We need to evaluate the polynomial at $x = -(-3)$ that is $x = 3$.
Replacing $x$ with $3$ in the given polynomial we get:
$4 \times 3^2 - 7 \times x - 15$
$= 4 \times 9 - 7 \times 3 - 15$
$= 36 - 21 - 15$
$= 36 - 36$
$= 0$
Therefore $x - 3$ is a factor of $4x^2 - 7x - 15$
Now let us check if $x + 4$ is a factor of $x^2 + x - 20$
Let us evaluate the given polynomial at $x = -4$
We get:
$(-4)^2 + (-4) - 20$
$= 16 - 4 - 20$
$= -8$
Therefore $(x + 4)$ is not a factor of $x^2 + x - 20$
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Now let us see if $(x - 4)$ is a factor of the same polynomial.
We evaluate the polynomial at $x = -(-4)$, that is, $x = 4$.
Replacing $x$ with $4$ we get:
$4^2 + 4 - 20$
$= 16 + 4 - 20$
$= 20 - 20$
$= 0$
Therefore, $(x-4)$ is a factor of $x^2 + x - 20$
This is valid for all higher order polynomials as well.