FACTORIZATION OF QUADRATIC EXPRESSIONS

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A quadratic expression is one in which the highest power of the unknown of the expression is $2$.

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 $y^2-64$

 $(y)^2-(8)^2=(y+8)(y-8)$


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