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# How do you add and simplify $3{{x}^{2}}-4x+7$ and $2{{x}^{3}}-4{{x}^{2}}-6x+5?$

Last updated date: 10th Aug 2024
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Hint: We use the polynomial addition to add the given second-degree polynomial and third-degree polynomial. We add the coefficients of the variables with the same power. Consider the polynomial with degree $2$ as the polynomial of degree $3$ with the coefficient of the variable with power $3$ is $0.$

Complete step by step solution:
Consider the given polynomials $3{{x}^{2}}-4x+7$ and $2{{x}^{3}}-4{{x}^{2}}-6x+5.$
When we are using the polynomial addition, we must remember that we can only add the coefficients of the variables with the same power.
Here, one polynomial is of degree $2$ and the other one is of degree $3.$
When we add these two polynomials, we add the coefficients of the similar terms to get the sum.
So, the sum will contain the coefficients which are the sum of the coefficients of the similar terms.
Let us keep the term ${{x}^{3}}$ aside for a while.
Now let us consider the term ${{x}^{2}}$ in both these polynomials.
The coefficient of the term ${{x}^{2}}$ in the polynomial of degree $2$ is $3$ and that in the polynomial of degree $3$ is $-4.$
Now we use the normal addition to add these coefficients to form the coefficient of the similar term in the sum polynomial.
So, we get $3+\left( -4 \right)=3-4=-1.$
Now consider the next term $x.$
The coefficient of $x$ in the polynomial of degree $2$ is $-4$ and that in the polynomial of degree $3$ is $-6.$
Using normal addition to get $-4+\left( -6 \right)=-4-6=-10.$
Now we consider the constant terms in both the polynomials.
The constant term in the second-degree polynomial is $7$ and the constant term in the third-degree polynomial is $5.$
Now we get $7+5=12.$
Since there is no term with the exponent $3$ in the second-degree polynomial we consider the coefficient of this term as zero. And the coefficient of the term ${{x}^{3}}$ in the third-degree polynomial is $2.$
We will get, $2+0=2.$
Hence, the simplified polynomial is $2{{x}^{3}}-{{x}^{2}}-10x+12.$

Note: This can be done with the following steps:
$\left( 3{{x}^{2}}-4x+7 \right)+\left( 2{{x}^{3}}-4{{x}^{2}}-6x+5 \right)=0{{x}^{3}}+3{{x}^{2}}-4x+7+2{{x}^{3}}-4{{x}^{2}}-6x+5$
That is, $\left( 3{{x}^{2}}-4x+7 \right)+\left( 2{{x}^{3}}-4{{x}^{2}}-6x+5 \right)=\left( 0+2 \right){{x}^{3}}+\left( 3-4 \right){{x}^{2}}+\left( -4-6 \right)x+\left( 7+5 \right)$
Hence, $\left( 3{{x}^{2}}-4x+7 \right)+\left( 2{{x}^{3}}-4{{x}^{2}}-6x+5 \right)=2{{x}^{3}}-{{x}^{2}}-10x+12.$