How do you simplify \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\]?
Answer
619.5k+ views
Hint: This type of problem can be solved with the help of distributive property. First, we have to consider the first two terms of the function, that is \[\left( x+6 \right)\left( x-8 \right)\]. Use distributive property, that is \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\]. Here, we find that a=x, b=6, c=x and d=-8. Substitute the values in the formula and expand the equation. Now, consider the whole function. We know that x-0=x. Thus we have to multiply x with all the terms of the simplified function.
Complete step by step solution:
According to the question, we are asked to simplify the function \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\].
We have been given the function \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\]. --------(1)
We first have to consider the first two terms of the function which is \[\left( x+6 \right)\left( x-8 \right)\].
We know that \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\]. We have to use this distributive property in the expression.
Here, we get a=x, b=6, c=x and d=-8.
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)=x\times x+x\times -8+6\times x+6\times -8\]
On further simplifications, we get
\[\left( x+6 \right)\left( x-8 \right)={{x}^{2}}-8x+6x-8\times 6\]
We know that the product of 6 and 8 is 48.
Therefore, we get
\[\left( x+6 \right)\left( x-8 \right)={{x}^{2}}-8x+6x-48\]
Now, we have to group all the x terms and add them.
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)={{x}^{2}}+\left( -8+6 \right)x-48\]
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)={{x}^{2}}-2x-48\]
Let us now substitute the simplified form in equation (1).
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)\left( x-0 \right)=\left( {{x}^{2}}-2x-48 \right)\left( x-0 \right)\]
We know that 0 subtracted from any term is equal to the term itself.
Therefore, we get
\[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)=\left( {{x}^{2}}-2x-48 \right)x\]
Let us now open the bracket and multiply x with all the terms.
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)\left( x-0 \right)={{x}^{2}}\times x-2x\times x-48x\]
On further simplification, we get
\[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)={{x}^{3}}-2{{x}^{2}}-48x\]
Therefore, the simplified form of \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\] is \[{{x}^{3}}-2{{x}^{2}}-48x\].
Note: Whenever we get such types of questions, we have to first consider the first two terms of the function. Solve the first and then multiply the simplified first two terms with the last term. This will avoid confusion and calculation mistakes. We can also solve this question by first considering the last two terms first, that is \[\left( x-8 \right)\left( x-0 \right)\]. On expanding this, we get \[{{x}^{2}}-8x\]. We then have to multiply \[{{x}^{2}}-8x\] with \[\left( x+6 \right)\] which is \[\left( x+6 \right)\left( {{x}^{2}}-8x \right)\] using the distributive property \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\].
Complete step by step solution:
According to the question, we are asked to simplify the function \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\].
We have been given the function \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\]. --------(1)
We first have to consider the first two terms of the function which is \[\left( x+6 \right)\left( x-8 \right)\].
We know that \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\]. We have to use this distributive property in the expression.
Here, we get a=x, b=6, c=x and d=-8.
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)=x\times x+x\times -8+6\times x+6\times -8\]
On further simplifications, we get
\[\left( x+6 \right)\left( x-8 \right)={{x}^{2}}-8x+6x-8\times 6\]
We know that the product of 6 and 8 is 48.
Therefore, we get
\[\left( x+6 \right)\left( x-8 \right)={{x}^{2}}-8x+6x-48\]
Now, we have to group all the x terms and add them.
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)={{x}^{2}}+\left( -8+6 \right)x-48\]
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)={{x}^{2}}-2x-48\]
Let us now substitute the simplified form in equation (1).
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)\left( x-0 \right)=\left( {{x}^{2}}-2x-48 \right)\left( x-0 \right)\]
We know that 0 subtracted from any term is equal to the term itself.
Therefore, we get
\[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)=\left( {{x}^{2}}-2x-48 \right)x\]
Let us now open the bracket and multiply x with all the terms.
\[\Rightarrow \left( x+6 \right)\left( x-8 \right)\left( x-0 \right)={{x}^{2}}\times x-2x\times x-48x\]
On further simplification, we get
\[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)={{x}^{3}}-2{{x}^{2}}-48x\]
Therefore, the simplified form of \[\left( x+6 \right)\left( x-8 \right)\left( x-0 \right)\] is \[{{x}^{3}}-2{{x}^{2}}-48x\].
Note: Whenever we get such types of questions, we have to first consider the first two terms of the function. Solve the first and then multiply the simplified first two terms with the last term. This will avoid confusion and calculation mistakes. We can also solve this question by first considering the last two terms first, that is \[\left( x-8 \right)\left( x-0 \right)\]. On expanding this, we get \[{{x}^{2}}-8x\]. We then have to multiply \[{{x}^{2}}-8x\] with \[\left( x+6 \right)\] which is \[\left( x+6 \right)\left( {{x}^{2}}-8x \right)\] using the distributive property \[\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd\].
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