Quadratic expressions are usually in the form $ax^2+bx+c$ where $a$, $b$ and $c$ are non-zero number.
For instance, $4x^2-3x+1$ is a quadratic expression as its highest power is $2$. Other quadratic expressions are $2x^2-4x-12$, $3x^2+15xy+10y^2$.
Some quadratic expression has factors. For example, $5x^2-13x-6$ has a factor of $(x-3)$ and $(5x+2)$. Just like $8\times4=32$, these factors can be multiplied to give $5x^2-13x-16$.
Others do not have a factor like $4x^2-3x-2$. Just like $17$ is said to be a prime number since it has no factor other than itself and one. Likewise, $4x^2-3x-2$ has no factor(other than itself and one).
To factorize a quadratic expression is to express it as a product of its factors. Accordingly, $5x^2-13x-6$ factories to $(x-3)(5x+2)$.
In the accompanying examples, you will learn the steps to follow when factoring quadratic expression.
Example 1
Factorize the quadratic expression $x^2+x-6$
The first step is to multiply the coefficient of the first and last term, that is $1\times-6=-6$
The listing term is to list all the pairs of a factor of $-6$ and their sum
Pairs of factor sum of factors
$(-1,6)$ $5$
$(-6,1)$ $-5$
$(-2,3)$ $1$
$(2,-3)$ $-1$
Out of all these factors, only $(-2,3)$ can replace the middle terms since they add up to the coefficient($1$) of the middle terms.
$x^2-2x+3x-6$
$(x^2-2x)(+3x-6)$
$x(x-2)+3(x-2)$ note: the factor inside the bracket should be the same
$(x+3)(x-2)$
Example 2
Factorize the quadratic expression $t^2+11t+18$
The first step is to multiply the coefficient of the first and last term, that is $1\times-18=18$
The listing term is to list all the pairs of a factor of $18$ and their sum
Pairs of factor sum of factors
$(1,18)$ $19$
$(2,9)$ $11$
$(3,6)$ $9$
Out of all these factors, only $(2,9)$ can replace the middle terms since they add up to the coefficient($11$) of the middle terms.
$t^2+2t+9t+18$
$(t^2+2t)(+9t+18)$
$t(t+2)+9(t+2)$
$(t+9)(t+2)$
Example 3
Factorize $3x^2-2x-5$ completely.
The first step is to multiply the coefficient of the first and last term, that is $3\times--5=-15$
The listing term is to list all the pairs of a factor of $-15$ and their sum
Pairs of factor sum of factors
$(-1,15)$ $14$
$(1,-15)$ $-14$
$(-3,5)$ $2$
$(3,-5)$ $-2$
As you can see, The only factor that can replace the middle number is $(3,-5)$ as they add up to the coefficient ($-2$) of the middle terms.
$3x^2+3x-5x-5$
$(3x^2+3x)(-5x-5)$
$3x(x+1)-5(x+1)$
$(3x-5)(x+1)$
Example 4
Factorize the expression $2y^2-3y-44$ completely
The first step is to multiply the coefficient of the first and last term, that is $2\times-44=-88$
The next step is to list all the pairs of a factor of $-88$ and their sum
Pairs of factor sum of factors
$(-1,88)$ $87$
$(1,-88)$ $-87$
$(-2,44)$ $42$
$(2,-44)$ $-42$
$(-4,22)$ $18$
$(4,-22)$ $-18$
$(-8,11)$ $3$
$(8,-11)$ $-3$
As you can see, The only factor that can replace the middle number is $(8,-11)$ as they add up to the coefficient ($-3$) of the middle terms.
$2y^2+8y-11y-44$
$2y(y+4)-11(y+4)$
$(2y-11)(y+4)$
Example 5
Factorise the expression $-2x^2+15+x$
First, let's rearrange the expression
$-2x^2+x+15$
To factorize this expression, the coefficient of $x^2$ need to be positive, hence we multiply the expression by $-1$
$-1(-2x^2+x+15)$
$2x^2-x-15$
When you multiply the first and last term, you will have $-30$
Here are the factor and their sums of $-30$
$(-1,30)$ $29$
$(1,-30)$ $-29$
$(-2,15)$ $13$
$(2,-15)$ $-13$
$(-3,10)$ $7$
$(3,-10)$ $-7$
$(-5,6)$ $1$
$(5,-6)$ $-1$
$(5,-6)$ adds up to $1$, hence, we used to replace $-x$
$2x^2-x-15$
$2x^2+5x-6x-15$
$(2x^2+5x)(-6x-15)$
$x(2x+5)-3(2x+5)$
$(x-3)(2x+5)$
Some quadratic expressions are the difference between two squares
Related post
Example 6
Factorize $y^2-64$
This is a difference of two square
$(y)^2-(8)^2=(y+8)(y-8)$
We have factorized quadratic expression. In this post, you will see some example that will solidify your knowledge of quadratic expression, do well to check it out here. Ok
If you got any questions, remember to ask our Facebook community.