Question

How do you multiply \[\left( 2x+2 \right)\left( x-4 \right)\]?

Answer 1

Hint: In this problem, we have to multiply the given factors to get an equation. We can multiply those factors in different methods, opening the brackets method, the other is using distributive property and using the FOIL method. In this problem, we can use opening the brackets method to multiply the factors.

Complete step-by-step solution:
We know that the given factors to be multiplied are,
\[\left( 2x+2 \right)\left( x-4 \right)\]
Now we can use opening the brackets method, that is open the brackets in the first factor and multiply the terms with the second factor.
\[\Rightarrow 2x\left( x-4 \right)+2\left( x-4 \right)\]
Now, we can multiply the terms outside the brackets to the terms inside the brackets, we get
\[\Rightarrow 2{{x}^{2}}-8x+2x-8\]
Now we can subtract the terms with x, we get
\[\Rightarrow 2{{x}^{2}}-6x-8\]
Therefore, by multiplying \[\left( 2x+2 \right)\left( x-4 \right)\] we get \[2{{x}^{2}}-6x-8\].

Note: We can also multiply the factors in the FOIL method to find the solution.
We know that the given factors to be multiplied are,
\[\left( 2x+2 \right)\left( x-4 \right)\]
We know that the term FOIL means Firsts Outsides Insides Lasts. We can apply this on the above factors we get,
Firsts \[2x\times x\Rightarrow 2{{x}^{2}}\]
Outsides \[2x\times -4\Rightarrow -8x\]
Insides \[2\times x\Rightarrow 2x\]
Lasts \[2\times -4\Rightarrow -8\]
Now we can add the above terms, we get
\[\Rightarrow 2{{x}^{2}}-8x+2x-8\]
Now we can subtract the terms with x, we get
\[\Rightarrow 2{{x}^{2}}-6x-8\]
Therefore, by multiplying the factors \[\left( 2x+2 \right)\left( x-4 \right)\] we get \[2{{x}^{2}}-6x-8\].
Students make mistakes in taking the correct terms for the FOIL method, which should be concentrated. We should know the multiplication table to solve these types of problems.