Question

Factorise the following expressions:
(1) \[2{{x}^{2}}+7x+3\]
(2) \[3{{x}^{2}}-x-4\]

Answer 1

Hint: We solve this problem by using the factorization method of quadratic equations.
If the quadratic expression is of the form then we use the factorisation method such that we rewrite the middle term as \[b=p+q\] such that \[p\times q=a\times c\] to get the common terms.
By using the above factorisation method we factorise the given statements one by one.

 Complete step by step answer:
We are given two expressions to factorise
Let us take the first expression
(1) \[2{{x}^{2}}+7x+3\]
Let us assume that the given expression as
\[\Rightarrow f\left( x \right)=2{{x}^{2}}+7x+3\]
Now, let us use the factorisation method to factorise the given expression
We know that if the quadratic expression is of the form \[a{{x}^{2}}+bx+c\]then we use the factorisation method such that we rewrite the middle term as \[b=p+q\] such that \[p\times q=a\times c\] to get the common terms.
By using the above statement let us take the two required equations as
\[\begin{align}
  & \Rightarrow 7=p+q \\
 & \Rightarrow 2\times 3=p\times q \\
\end{align}\]
Now, by trial and error method we get the value of \[p,q\] as 6, 1 respectively.
By rewriting the given expression by using the factorisation method then we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=2{{x}^{2}}+6x+x+3 \\
 & \Rightarrow f\left( x \right)=2x\left( x+3 \right)+1\left( x+3 \right) \\
 & \Rightarrow f\left( x \right)=\left( x+3 \right)\left( 2x+1 \right) \\
\end{align}\]
Therefore we can conclude that the factorisation of given expression as
\[\therefore 2{{x}^{2}}+7x+3=\left( x+3 \right)\left( 2x+1 \right)\]
Now, let us take the second expression that is
(2) \[3{{x}^{2}}-x-4\]
Let us assume that the given expression as
\[\Rightarrow f\left( x \right)=3{{x}^{2}}-x-4\]
Now, let us use the factorisation method to factorise the given expression
We know that if the quadratic expression is of the form \[a{{x}^{2}}+bx+c\]then we use the factorisation method such that we rewrite the middle term as \[b=p+q\] such that \[p\times q=a\times c\] to get the common terms.
By using the above statement let us take the two required equations as
\[\begin{align}
  & \Rightarrow -1=p+q \\
 & \Rightarrow \left( 3 \right)\times \left( -4 \right)=p\times q \\
\end{align}\]
Now, by trial and error method we get the value of \[p,q\] as -4, 3 respectively.
By rewriting the given expression by using the factorisation method then we get
\[\begin{align}
  & \Rightarrow f\left( x \right)=3{{x}^{2}}+3x-4x-4 \\
 & \Rightarrow f\left( x \right)=3x\left( x+1 \right)-4\left( x+1 \right) \\
 & \Rightarrow f\left( x \right)=\left( x+1 \right)\left( 3x-4 \right) \\
\end{align}\]
Therefore we can conclude that the factorisation of given expression as
\[\therefore 3{{x}^{2}}-x-4=\left( x+1 \right)\left( 3x-4 \right)\]

Note:
We need to note that we can find the value of \[p,q\] in the factorisation method from trial and error method because they are integers.
Generally we find the value of variables knowing the value of sum and product by using the algebra formula that is
\[{{\left( p-q \right)}^{2}}={{\left( p+q \right)}^{2}}-4pq\]
Then we solve for \[p,q\]
But in the process of factorisation of quadratic expression, we can use the trial and error method as they are integers. We use the above-mentioned formula when \[p,q\] are real which includes rational and irrational also.