
How do you multiply $ \left( 3x+1 \right)\left( x-2 \right) $ ?
Answer
559.5k+ views
Hint: In the problem, we need to multiply the given two terms. Here we can follow the FOIL rule to multiply the given terms. So, we will first multiply the individual terms according to the FOIL rule which is nothing but First, Outside, Inner, Last terms. After multiplying the term individually, we will add all the calculated values to get the required result.
Complete step by step answer:
Given that, $ \left( 3x+1 \right)\left( x-2 \right) $ .
First terms in the given equation are $ 3x $ , $ x $ . Multiplying the both the terms, then we will get
$ 3x\times x=3{{x}^{2}}...\left( \text{i} \right) $
Outside terms in the given equation are $ 3x $ , $ -2 $ . Multiplying the both the terms, then we will get
$ 3x\times -2=-6x...\left( \text{ii} \right) $
Inner terms in the given equation are $ 1 $ , $ x $ . Multiplying the both the terms, then we will get
$\Rightarrow$ $ 1\times x=x....\left( \text{iii} \right) $
Last terms in the given equation are $ 1 $ , $ -2 $ . Multiplying the both the terms, then we will get
$\Rightarrow$ $ 1\times -2=-2....\left( \text{iv} \right) $
Now the product of the given equation is the algebraic sum of the products obtained by the FOIL rule. From equations $ \left( \text{i} \right) $ , $ \left( \text{ii} \right) $ , $ \left( \text{iii} \right) $ , $ \left( \text{iv} \right) $ we can write
$\Rightarrow$ $ \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-6x+x-2 $
Simplifying the above equation, then we will get
$ \Rightarrow \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-5x-2 $
Hence the product of the equation $ \left( 3x+1 \right)\left( x-2 \right) $ is $ 3{{x}^{2}}-5x-2 $ .
Note:
We can directly multiply the two terms and use the distribution law of multiplication over the addition and subtraction, then we will get the required result.
Given $ \left( 3x+1 \right)\left( x-2 \right) $ .
Multiplying each term individually, then we will get
$ \left( 3x+1 \right)\left( x-2 \right)=3x\left( x-2 \right)+1\left( x-2 \right) $
Using the distribution law of multiplication over the subtraction, then we will get
$ \begin{align}
& \left( 3x+1 \right)\left( x-2 \right)=3x\times x-3x\times 2+1\times x-1\times 2 \\
& \Rightarrow \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-6x+x-2 \\
& \Rightarrow \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-5x+2 \\
\end{align} $
From both the methods we got the same result.
Complete step by step answer:
Given that, $ \left( 3x+1 \right)\left( x-2 \right) $ .
First terms in the given equation are $ 3x $ , $ x $ . Multiplying the both the terms, then we will get
$ 3x\times x=3{{x}^{2}}...\left( \text{i} \right) $
Outside terms in the given equation are $ 3x $ , $ -2 $ . Multiplying the both the terms, then we will get
$ 3x\times -2=-6x...\left( \text{ii} \right) $
Inner terms in the given equation are $ 1 $ , $ x $ . Multiplying the both the terms, then we will get
$\Rightarrow$ $ 1\times x=x....\left( \text{iii} \right) $
Last terms in the given equation are $ 1 $ , $ -2 $ . Multiplying the both the terms, then we will get
$\Rightarrow$ $ 1\times -2=-2....\left( \text{iv} \right) $
Now the product of the given equation is the algebraic sum of the products obtained by the FOIL rule. From equations $ \left( \text{i} \right) $ , $ \left( \text{ii} \right) $ , $ \left( \text{iii} \right) $ , $ \left( \text{iv} \right) $ we can write
$\Rightarrow$ $ \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-6x+x-2 $
Simplifying the above equation, then we will get
$ \Rightarrow \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-5x-2 $
Hence the product of the equation $ \left( 3x+1 \right)\left( x-2 \right) $ is $ 3{{x}^{2}}-5x-2 $ .
Note:
We can directly multiply the two terms and use the distribution law of multiplication over the addition and subtraction, then we will get the required result.
Given $ \left( 3x+1 \right)\left( x-2 \right) $ .
Multiplying each term individually, then we will get
$ \left( 3x+1 \right)\left( x-2 \right)=3x\left( x-2 \right)+1\left( x-2 \right) $
Using the distribution law of multiplication over the subtraction, then we will get
$ \begin{align}
& \left( 3x+1 \right)\left( x-2 \right)=3x\times x-3x\times 2+1\times x-1\times 2 \\
& \Rightarrow \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-6x+x-2 \\
& \Rightarrow \left( 3x+1 \right)\left( x-2 \right)=3{{x}^{2}}-5x+2 \\
\end{align} $
From both the methods we got the same result.
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