Question

How do you simplify $\dfrac{2{{x}^{2}}+5x+2}{4{{x}^{2}}-1}\times \dfrac{2{{x}^{2}}+x-1}{{{x}^{2}}+x-2}?$

Answer 1

Hint: We will factorize each of the polynomials in the given expression. That is, we will factorize the polynomials in the numerators and the denominators of both fractions. Then we will cancel the greatest common factor from both the fractions. Since the operation is multiplication, the operation is directly done on the numerators of both fractions and the denominators of both fractions. We can generate a single fraction where the numerator contains the product of numerators of both fractions and the denominator contains the product of both denominators.

Complete step by step solution:
Let us consider the given expression to be simplified, $\dfrac{2{{x}^{2}}+5x+2}{4{{x}^{2}}-1}\times \dfrac{2{{x}^{2}}+x-1}{{{x}^{2}}+x-2}.$
We can easily simplify this expression by factoring each of the polynomials.
Let us consider the polynomial in the numerator of the first fraction, $2{{x}^{2}}+5x+2.$
Now, we can change this polynomial as $2{{x}^{2}}+5x+2=2{{x}^{2}}+4x+x+2$ by splitting the middle term.
We will get $2{{x}^{2}}+5x+2=2x\left( x+2 \right)+x+2.$
From this, we will get $2{{x}^{2}}+5x+2=\left( 2x+1 \right)\left( x+2 \right).$
We know the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}.$
So, the denominator of the first fraction can be factored to $4{{x}^{2}}-1=\left( 2x+1 \right)\left( 2x-1 \right).$
We can factorise the polynomial $2{{x}^{2}}+x-1$ into $\left( 2x-1 \right)\left( x+1 \right)$ by adding and subtracting $x$ as $2{{x}^{2}}+x-1=2{{x}^{2}}+2x-x-1=2x\left( x+1 \right)-\left( x+1 \right).$
We will factorise the polynomial ${{x}^{2}}+x-2$ into $\left( x+2 \right)\left( x-1 \right)$ by adding and subtracting $x$ as ${{x}^{2}}+x-2={{x}^{2}}+2x-x-2=x\left( x+2 \right)-\left( x+2 \right).$
Let us substitute for the polynomials in the given expression.
We will get $\dfrac{2{{x}^{2}}+5x+2}{4{{x}^{2}}-1}\times \dfrac{2{{x}^{2}}+x-1}{{{x}^{2}}+x-2}=\dfrac{\left( 2x+1 \right)\left( x+2 \right)}{\left( 2x+1 \right)\left( 2x-1 \right)}\times \dfrac{\left( 2x-1 \right)\left( x+1 \right)}{\left( x+2 \right)\left( x-1 \right)}.$
Let us cancel the common factor \[2x+1\] to get \[\dfrac{\left( x+2 \right)}{\left( 2x-1 \right)}\times \dfrac{\left( 2x-1 \right)\left( x+1 \right)}{\left( x+2 \right)\left( x-1 \right)}.\]
Next, we will cancel $x+2$ to get \[\dfrac{1}{\left( 2x-1 \right)}\times \dfrac{\left( 2x-1 \right)\left( x+1 \right)}{\left( x-1 \right)}.\]
We can cancel $2x-1$ also to get \[\dfrac{1}{1}\times \dfrac{\left( x+1 \right)}{\left( x-1 \right)}=\dfrac{\left( x+1 \right)}{\left( x-1 \right)}.\]

Hence the simplified form of the given expression is \[\dfrac{\left( x+1 \right)}{\left( x-1 \right)}.\]

Note: Since the operation is multiplication, we can multiplicate the numerators directly and similarly we can multiply the denominators directly to find the product. Also, we can divide the numerator of first fraction with the denominator of second fraction and we can divide the numerator of second fraction with the denominator of first fraction.