Question

How do you add the expression $\left( 4{{v}^{3}}-3{{v}^{2}}+2v \right)+\left( 3{{v}^{3}}+6{{v}^{2}}-9v \right)?$

Answer 1

Hint: We do the polynomial addition. What we are going to do here is to add the coefficients of the corresponding terms. The sum of polynomials will contain the terms with coefficients which are sums of the corresponding coefficients of the two summand polynomials.

Complete step by step solution:
Consider the given sum $\left( 4{{v}^{3}}-3{{v}^{2}}+2v \right)+\left( 3{{v}^{3}}+6{{v}^{2}}-9v \right).$
In this sum, the two summands $\left( 4{{v}^{3}}-3{{v}^{2}}+2v \right)$ and $\left( 3{{v}^{3}}+6{{v}^{2}}-9v \right)$ are two polynomials of degree three with the constant terms equal to zero.
To find the sum of these two polynomials we use the polynomial addition in which we add only the coefficients of the corresponding terms.
By corresponding terms, we mean the variables with the same power.
That is, ${{v}^{3}}$ from the first summand and ${{v}^{3}}$ from the second summand.
So, we add the coefficient of ${{v}^{3}}$ from the first summand and the coefficient of ${{v}^{3}}$ from the second summand to get the coefficient of ${{v}^{3}}$ in the sum polynomial.
The coefficient of ${{v}^{3}}$ in the first summand is $4$ and the coefficient of ${{v}^{3}}$ in the second summand is $3.$
Therefore, the coefficient of ${{v}^{3}}$ in the sum polynomial is $4+3=7.$
Similarly, when we consider the second terms of both the summand polynomials which is ${{v}^{2}},$ the coefficients of ${{v}^{2}}$ from the first and the second summands are $-3$ and $6$ respectively.
We add them together to get the coefficient of ${{v}^{2}}$ in the sum polynomial.
Therefore, the coefficient of the second term ${{v}^{2}}$ in the sum polynomial is $-3+6=3.$
We repeat the same procedure with the coefficients of the last terms $v$ from both the summand polynomials.
The coefficients of $v$ in the first and the second summands are $2$ and $-9.$
Therefore, the coefficient of the term $v$ in the sum polynomial is $2+\left( -9 \right)=2-9=-7.$
Now, we get the sum polynomial as $7{{v}^{3}}+3{{v}^{2}}-7v.$
Hence, $\left( 4{{v}^{3}}-3{{v}^{2}}+2v \right)+\left( 3{{v}^{3}}+6{{v}^{2}}-9v \right)=7{{v}^{3}}+3{{v}^{2}}-7v.$

Note: We use normal addition to add the coefficients. The above procedure can be written in two steps as $\left( 4{{v}^{3}}-3{{v}^{2}}+2v \right)+\left( 3{{v}^{3}}+6{{v}^{2}}-9v \right)=\left( 4+3 \right){{v}^{3}}+\left( -3+6 \right){{v}^{2}}+\left( 2-9 \right)v$
That is, $\left( 4{{v}^{3}}-3{{v}^{2}}+2v \right)+\left( 3{{v}^{3}}+6{{v}^{2}}-9v \right)=7{{v}^{3}}+3{{v}^{2}}-7v.$