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Factoring by grouping


This article covers in detail, the techniques that can be used for solving quadratic equations by factoring.
There can be three types of quadratic functions depending on the number of terms. when coming to factoring the quadratics, there are two key types of quadratic equations of interest they are

  1. Quadratic Binomial
  2. Quadratic Trinomial.

The approach or the method of factorization differs by the type of a quadratic function.
Following are the key features of this article and please do attempt the quiz at the end, to check your understanding:

  • A brief introduction on what does factoring of a quadratic mean.
  • When can we use the factorization method?
  • The role of discriminant in solving quadratic equations by factoring
  • Factoring quadratic binomial.
  • Factoring quadratic trinomial and how to factor by grouping.
  • Worked out Examples

1.Solving quadratic equations by factoring:


i) What is factoring the quadratic equation?

Factoring quadratic is an approach to find the roots of a quadratic equation.

Solving quadratic equations by factoring is all about writing the quadratic function as a product of two binomials functions of one degree each.

For example the trinomail quadratic   x^{2}+8x +12=0 ,can we written as  (x+6)(x+2)=0, where (x+2) and (x+6) are the binomial terms each of degree 1.

Further, we can say x+6=0 and x+2=0 and x =-6,-2 thereby are the roots

Once we find the two binomials we can find the solutions/roots of the quadratic easy thereby.

ii) The solutions to the quadratic equation are the X-intercepts:

In fact, solving quadratic equations by factoring is none other than finding the X-intercepts, which are the points where the graph of the quadratic equation intersects the “x” axis.

If the graph of the quadratic equation intersects the X-axis at least one point, then we say that there are “Real Roots” for a given quadratic.

If a graph of a quadratic equation does not intersect “x” axis, then we say that there are no “REAL roots” available for the given quadratic equation.

These roots are the “complex roots”.So if a quadratic equation has “COMPLEX roots” it does not intersect the “x” axis.

Since they are points on “x”-axis, “y” coordinate will be zero. In general, the point looks like (x,0).


iii) When can we use Factorisation method? 

Solving quadratic equations by factoring at times is tedious.

The interesting fact is that we can decide upfront, whether factorisation method can be used or not, using the discriminant of a quadratic equation.

Let’s us look how it works and concept behind it.

For factoring a  quadratic equation, there must be  Real and Rational roots available for sure.

In other words, if a quadratic equation has “COMPLEX roots” or “IR-Rational Roots”, it  “CANNOT” be solved by factoring.

But how do we know whether a given quadratic equation has “REAL roots” Or “IR-Rational Roots” Or”COMPLEX roots”? Is there any way to identify?

The answer is yes!.

Before applying factorization method, we need to inspect the discriminant of the quadratic equation  y= ax^{2}+bx +c .


iv)The role of discriminant in solving quadratic equations by factoring

we can go ahead Solving quadratic equations by factoring, by closely examining the discriminant of a quadratic.

The discriminant of a quadratic is given by an expression   b^{2}-4ac . If and only if the Discriminant is a perfect square like( 4,9,16,25..) the quadratic equation can be solved by factoring.

If the Discriminant is <0 or not a perfect square, the quadratic equation cannot be factorized.

In such scenarios, we need to use alternate methods like the “Quadratic Formula” to get the IR-Rational Roots or Complex roots of the quadratic.


2. Quadratic Binomial

i)Quadratic Binomial definition:

A quadratic binomial is a quadratic function containing only two terms.

The term “binomial” conveys the presence of  2 terms.For example 2x +1, x^{2}+5 have only two terms in them

The term “Quadratic” which adds upon conveys that the highest degree must be two.

Therefore the term “Quadratic Binomial” means a quadratic function with the highest order as 2 and the maximum number of terms as two.


ii)Factoring Quadratic Binomial:

Solving a quadratic binomial is comparatively easier than solving a quadratic trinomial.

All that needed is to pull out the common factor and get the solutions.

There can be two types of quadratic binomials:-

1.x^{2} with the constant term. example x^{2}-5 =0

2. x^{2} with a “x” term example x^{2}+3x=0


Example1: Find the roots of the Quadratic binomial  x^{2}+3x=0

Solution: Factoring a Quadratic Binomial is done by pulling out a common factor of the two available terms.

Step1: Take the common factor out of the two available terms.

In x^{2}+3x=0 , we have  a  common factor “x”.Therefore take  “x” as a common factor we have \begin{aligned}x^{2}+3x=0\\ \Rightarrow x\left( x+3\right) =0\end{aligned}

Step2: Using the zero multiplicity property, solve for “x”.

Using the property  we get  x =0 , x+3=0.Therefore the solutions of x are 0 and -3.


Example 2: Find the roots of the Quadratic binomial  2x^{2}-4x=0

Solution: In the given problem we have the common factor 2x.therefore we need to take 2x as the common factor.

\begin{aligned}2x^{2}-4x=0\\ \Rightarrow 2x\left( x-4\right) =0\\ \Rightarrow 2x=0,x-4=0\\ \Rightarrow x=0,x=4\end{aligned}


3. Quadratic Trinomial

i)Quadratic Trinomial definition:

A Quadratic Trinomial has all the three terms. The x^2, x and the constant term.

As we are finding the roots, the points on “x” axis, the first step to be performed is to replace the “y” with “0” in the Quadratic equation   y= ax^{2}+bx +c .

In  ax^{2}+bx +c=0 ,we need t o pull the common factor out and simplify the expression.

The middle term with “x” and coefficient  “b” to be split into two parts say p and q  such that, the sum of the two parts p+q= b and product of the two parts p*q is the product of a&c.

p+q must be equal “b” and at the same time, p*q must be equal to a*c.

At times it would be tedious to make a correct split of the coefficient “b” into two parts.

An easy approach is, to write a*c as the “The product of prime factors”.

Once we get the prime factor using tree method, we can identify a combination which when added or subtracted would yield the coefficient “b”.

Therefore .the key job in solving quadratic equations by factoring is, identifying the correct split of the middle term.

Let’s view an example which implements the above method.


ii)How to Split of the coefficient “b” for the factorization using the product of prime factors?:-

For example, let’s say we have  y= 4x^{2}+10x +6 .

On comparision we have the coefficients a= 4, b= 10 and  c=6 therefore a*c=4(6) =24.

The coefficient “b”  is to be split into two parts p and q, such that p+q  must be equal to 10 and p*q must be equal  24.

Writing the 24 as the product of prime factors we have 24=2*2*2*3.

Now we need to choose the values of p and q from the combination of prime factors.

Let us take “p” as  2*2*3 =12 and “q” as 2.These two numbers 12 and 2 can yield a 10.

Choosing p as 12 and q as -2 we will have p+q =12+(-2)=10 which is the “b”.

But unfortunately p*q=12 *(-2)= -24 which we are not looking for -24. we need p*q = 24.

Therefore P= 12 and the q=-2 combination does not work.

Let us now choose a little-balanced combination for “p” as   2*2 =4 and “q” as 2*3 =6.

These two numbers 6 and 4 can yield a 10 .p+q= 6+4 =10 which is “b”.

And also p*q=4*6=24 as desired. Hence the correct split for the coefficient “b” is p=6 and q=4.


iii) Factor by grouping 

Once the correct split of the middle terms is made, we can group the terms and factor them up for example. 4x^{2}+10x +6 can be written as  4x^{2}+6x + 4x +6 .

we can group the first two terms   4x^{2}+6x and pull the greatest common factor out. We observe that 2x is the greatest common factor of the first two terms and 2 is the Greatest common factor of 4x  and 6 the next two terms.

 4x^{2}+6x + 4x +6 \\\begin{aligned}\left( 4x^{2}+6x\right) +\left( 4x+6\right) =0\\ \Rightarrow 2x\left( 2x+3\right) +2\left( 2x+3\right) =0\\ \Rightarrow \left( 2x+3\right) \left( 2x+2\right) =0\end{aligned}.


4.Worked out examples

Find the roots of given quadratic equation.
i) y=4x^{2}-6x -10

Strategy Step-by-Step:-

 Step1: Replace y with 0 as we are finding “x” intercepts.Identify the coefficients a,b,c of the quadratic equation.

  • In the current problem,replacing y with 0, we have 0=4x^{2}-6x -10.On comparison with the standard form the quadratic equation, y= ax^{2}+bx +c we can conclude that a=4 , b= -6 and c=-10.


Step2: Find the Discriminant   b^{2}-4ac .  

If the discriminant is a perfect square, then we can proceed further with factorization.

  • The discriminant  b^{2}-4ac   would be  (-6)^{2}-4*4*(-10) .So the discriminant is 36-(-160) = 36+160 = 196. Since 196 =  14^{2} a perfect square,We can go ahead with factorisation of the quadratic given.


Step3: Find the common factors, if any, of the three terms a,b,c.

Pull the common factor out to simplify the quadratic equation.

  • The coefficients  4,-6,1-0 have a common factor  2.So the given quadratic equation  4x^{2}-6x -10 can we rewritten as  2(2x^{2}-3x -5)=0 .

Therefore the simplified equation is  2x^{2}-3x -5=0 .


Step4: Split the Coefficient  “b” into two parts say p and q  such that, the sum of the two parts p+q= b and product of the two parts p*q is the product of a&c.rewite the quadratic equation with the split terms.

  • We found the value of b as -3 and the value of a*c would be 2*(-5) = -10.

The correct split would be p=-5 and q=2.Therefore   2x^{2}-3x -5=0 can be written as    2x^{2}-5x  +2x-5=0 .

Splitting the middle term -3x as -5x +2x


Step5: Factor by grouping by taking Greatest Common Factor(G.C.F) out of the first two terms and next two terms

  • In the  given equation we  can observe that the  GCF of the  first  two  terms  2x^{2}-5x is “x”   and the  next  two  terms   +2x -5 is 1.Therefore we can write  x(2x-5) and 1(2x-5).


Step6: Take out the G.C.F of the terms inturn to write the Final factorized expression.

  • We, in turn, observe that  (2x-5) is the common factor available in the two terms.

Hence the final factorized expression can be  written as  (2x-5)(x+1)=0 .


Step7: Using zero multiplication property, find the zeros of the function.

5). Got a troubling math question?confused emoji

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