
How do you combine like terms in $5c\left( 2{{c}^{2}}-3c+4 \right)+2c\left( 7c-8 \right)$?
Answer
537.9k+ views
Hint: Firstly, we need to simplify the brackets in the polynomial given in the above question. For this, we have to use the distributive law of the algebraic multiplication, which is given as $a\left( b+c+d \right)=ab+bc+ad$. By applying this law on the given polynomial, we will be able to simplify both the brackets and will obtain the polynomial as the sum of terms of different powers of c. The like terms in a polynomial are the terms having the same exponent of the variable. In the polynomial, the different powers on c are from one to three.
Complete step by step solution:
Let us consider the polynomial given in the above question as
$\Rightarrow f\left( c \right)=5c\left( 2{{c}^{2}}-3c+4 \right)+2c\left( 7c-8 \right)$
Using the distributive law of the algebraic multiplication, given as $a\left( b+c+d \right)=ab+bc+ad$, we can simplify the two brackets of the above polynomial as
\[\begin{align}
& \Rightarrow f\left( c \right)=5c\left( 2{{c}^{2}} \right)+5c\left( -3c \right)+5c\left( 4 \right)+2c\left( 7c \right)+2c\left( -8 \right) \\
& \Rightarrow f\left( c \right)=10{{c}^{3}}-15{{c}^{2}}+20c+14{{c}^{2}}-16c \\
\end{align}\]
Now, we know that the like terms in a polynomial are the terms having the equal exponent on the variable. We can see that in the above polynomial, the different powers are three, two, and one. In order to combine the like terms, we first arrange them together to get
$\Rightarrow f\left( c \right)=10{{c}^{3}}-15{{c}^{2}}+14{{c}^{2}}+20c-16c$
Now, we can add the coefficients of the like terms of the above polynomial to finally get
\[\begin{align}
& \Rightarrow f\left( c \right)=10{{c}^{3}}+\left( -15+14 \right){{c}^{2}}+\left( 20-16 \right)c \\
& \Rightarrow f\left( c \right)=10{{c}^{3}}+\left( -1 \right){{c}^{2}}+\left( 4 \right)c \\
& \Rightarrow f\left( c \right)=10{{c}^{3}}-{{c}^{2}}+4c \\
\end{align}\]
Hence, we have combined the like terms of the given polynomial and finally got it as \[10{{c}^{3}}-{{c}^{2}}+4c\].
Note: We must make sure that the terms are arranged in the decreasing order of the powers of c in the final obtained expression. This is the conventional way of representing a polynomial. Also, while simplifying the brackets, we must take care of the signs of different terms.
Complete step by step solution:
Let us consider the polynomial given in the above question as
$\Rightarrow f\left( c \right)=5c\left( 2{{c}^{2}}-3c+4 \right)+2c\left( 7c-8 \right)$
Using the distributive law of the algebraic multiplication, given as $a\left( b+c+d \right)=ab+bc+ad$, we can simplify the two brackets of the above polynomial as
\[\begin{align}
& \Rightarrow f\left( c \right)=5c\left( 2{{c}^{2}} \right)+5c\left( -3c \right)+5c\left( 4 \right)+2c\left( 7c \right)+2c\left( -8 \right) \\
& \Rightarrow f\left( c \right)=10{{c}^{3}}-15{{c}^{2}}+20c+14{{c}^{2}}-16c \\
\end{align}\]
Now, we know that the like terms in a polynomial are the terms having the equal exponent on the variable. We can see that in the above polynomial, the different powers are three, two, and one. In order to combine the like terms, we first arrange them together to get
$\Rightarrow f\left( c \right)=10{{c}^{3}}-15{{c}^{2}}+14{{c}^{2}}+20c-16c$
Now, we can add the coefficients of the like terms of the above polynomial to finally get
\[\begin{align}
& \Rightarrow f\left( c \right)=10{{c}^{3}}+\left( -15+14 \right){{c}^{2}}+\left( 20-16 \right)c \\
& \Rightarrow f\left( c \right)=10{{c}^{3}}+\left( -1 \right){{c}^{2}}+\left( 4 \right)c \\
& \Rightarrow f\left( c \right)=10{{c}^{3}}-{{c}^{2}}+4c \\
\end{align}\]
Hence, we have combined the like terms of the given polynomial and finally got it as \[10{{c}^{3}}-{{c}^{2}}+4c\].
Note: We must make sure that the terms are arranged in the decreasing order of the powers of c in the final obtained expression. This is the conventional way of representing a polynomial. Also, while simplifying the brackets, we must take care of the signs of different terms.
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