How do you factor completely: \[4{{x}^{2}}+16x\]?
Answer
624.3k+ views
Hint: To factorize the equation \[4{{x}^{2}}+16x\] completely, we have to assume the given equation with a variable and consider it as an equation. We have to pull the terms 4, x from the equation by then we will get three factors and therefore we can factorize the equation as well as we can find the factors for the equation.
Complete step-by-step solution:
For the given problem we are given to factor completely for the equation\[4{{x}^{2}}+16x\].
So, for solving this question let us assume the equation as ‘A’ and consider the above equation as equation (1).
Let us consider
\[A=4{{x}^{2}}+16x.......\left( 1 \right)\]
By observing the equation (1) we can say that there is no constant term in this equation (1).
So therefore it is very easy to solve the given equation.
We can clearly see that 4 is common in both the terms so let us pull it out from the equation. For that we have to write 16 as 4.4.
Rewriting the equation (1) by replacing 16 with 4.4, we get
\[\Rightarrow A=4{{x}^{2}}+4.4x\]
Let us consider the above equation as equation (2).
\[A=4{{x}^{2}}+4.4x...........\left( 2 \right)\]
So let us pull 4 from the equation, we get
\[\Rightarrow A=4.\left( {{x}^{2}}+4x \right)\]
Let us consider the above equation as equation (3).
\[\Rightarrow A=4\left( {{x}^{2}}+4x \right)...........\left( 3 \right)\]
Now let us take x from the equation, we get
\[\Rightarrow A=\left( 4 \right)\left( x \right)\left( x+4 \right)\]
Let us consider the above equation as equation (4).
\[\Rightarrow A=\left( 4 \right)\left( x \right)\left( x+4 \right)..........\left( 4 \right)\]
Therefore, we can say that there is no chance to take anything from the equation (4).
Hence by the equation (4) we can say that \[\left( \text{4} \right)\text{,}\left( \text{x} \right)\text{and}\left( \text{x+4} \right)\] are the factors of equation (1).
Note: If the problem is given with a constant then we factorize the equation using the middle term expansion. If the examiner asks to find factors for the same question we have to specify our factors as I specified in the last line of the solution. Students should mind calculation mistakes.
Complete step-by-step solution:
For the given problem we are given to factor completely for the equation\[4{{x}^{2}}+16x\].
So, for solving this question let us assume the equation as ‘A’ and consider the above equation as equation (1).
Let us consider
\[A=4{{x}^{2}}+16x.......\left( 1 \right)\]
By observing the equation (1) we can say that there is no constant term in this equation (1).
So therefore it is very easy to solve the given equation.
We can clearly see that 4 is common in both the terms so let us pull it out from the equation. For that we have to write 16 as 4.4.
Rewriting the equation (1) by replacing 16 with 4.4, we get
\[\Rightarrow A=4{{x}^{2}}+4.4x\]
Let us consider the above equation as equation (2).
\[A=4{{x}^{2}}+4.4x...........\left( 2 \right)\]
So let us pull 4 from the equation, we get
\[\Rightarrow A=4.\left( {{x}^{2}}+4x \right)\]
Let us consider the above equation as equation (3).
\[\Rightarrow A=4\left( {{x}^{2}}+4x \right)...........\left( 3 \right)\]
Now let us take x from the equation, we get
\[\Rightarrow A=\left( 4 \right)\left( x \right)\left( x+4 \right)\]
Let us consider the above equation as equation (4).
\[\Rightarrow A=\left( 4 \right)\left( x \right)\left( x+4 \right)..........\left( 4 \right)\]
Therefore, we can say that there is no chance to take anything from the equation (4).
Hence by the equation (4) we can say that \[\left( \text{4} \right)\text{,}\left( \text{x} \right)\text{and}\left( \text{x+4} \right)\] are the factors of equation (1).
Note: If the problem is given with a constant then we factorize the equation using the middle term expansion. If the examiner asks to find factors for the same question we have to specify our factors as I specified in the last line of the solution. Students should mind calculation mistakes.
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