
How do you factor the expression \[{{x}^{2}}+6x+8\]?
Answer
529.5k+ views
Hint: In this problem, we have to find the factor of the given expression by factorization method. We can first split the middle term, i.e. the x term with its coefficient. We have to expand the middle term in such a way that their addition is equal to the middle term i.e. 6x, and multiplication is equal to \[4\times 2=8\], where \[4+2=6\] which is the middle term. We can then take common terms outside to get the factors.
Complete step-by-step solution:
We know that the given expression is,
\[{{x}^{2}}+6x+8\]
We can first split the middle term to form factors.
We have to expand the middle term in such a way that their addition is equal to the middle term i.e.6x, and multiplication is equal to \[4\times 2=8\], where \[4+2=6\] which is the middle term.
\[\Rightarrow {{x}^{2}}+4x+2x+8\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( {{x}^{2}}+4x \right)+\left( 2x+8 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow x\left( x+4 \right)+2\left( x+4 \right)\]
We can again take the common factor first then the remaining terms, to make a factor, we get
\[\Rightarrow \left( x+2 \right)\left( x+4 \right)\]
Therefore, the factors are \[\left( x+2 \right)\left( x+4 \right)\].
Note: We can also verify for the correct answer using the quadratic formula.
The quadratic formula for the equation \[a{{x}^{2}}+bx+c\]is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We know that the given expression is,
\[{{x}^{2}}+6x+8\]
Where, a = 1, b = 6, c = 8.
We can substitute the above values in quadratic formula.
\[\begin{align}
& \Rightarrow x=\dfrac{-1\pm \sqrt{36-32}}{2} \\
& \Rightarrow x=\dfrac{-1\pm \sqrt{4}}{2} \\
& \Rightarrow x=\dfrac{-1\pm 2}{2} \\
& \Rightarrow x=-4,-2 \\
\end{align}\]
Where,\[x=-2,x=-4\]
We can take the term in the right-hand side, to the left-hand side by changing its sign.
\[\Rightarrow \left( x+2 \right)\left( x+4 \right)\]
Therefore, the factors are \[\left( x+2 \right)\left( x+4 \right)\].
Complete step-by-step solution:
We know that the given expression is,
\[{{x}^{2}}+6x+8\]
We can first split the middle term to form factors.
We have to expand the middle term in such a way that their addition is equal to the middle term i.e.6x, and multiplication is equal to \[4\times 2=8\], where \[4+2=6\] which is the middle term.
\[\Rightarrow {{x}^{2}}+4x+2x+8\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( {{x}^{2}}+4x \right)+\left( 2x+8 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow x\left( x+4 \right)+2\left( x+4 \right)\]
We can again take the common factor first then the remaining terms, to make a factor, we get
\[\Rightarrow \left( x+2 \right)\left( x+4 \right)\]
Therefore, the factors are \[\left( x+2 \right)\left( x+4 \right)\].
Note: We can also verify for the correct answer using the quadratic formula.
The quadratic formula for the equation \[a{{x}^{2}}+bx+c\]is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We know that the given expression is,
\[{{x}^{2}}+6x+8\]
Where, a = 1, b = 6, c = 8.
We can substitute the above values in quadratic formula.
\[\begin{align}
& \Rightarrow x=\dfrac{-1\pm \sqrt{36-32}}{2} \\
& \Rightarrow x=\dfrac{-1\pm \sqrt{4}}{2} \\
& \Rightarrow x=\dfrac{-1\pm 2}{2} \\
& \Rightarrow x=-4,-2 \\
\end{align}\]
Where,\[x=-2,x=-4\]
We can take the term in the right-hand side, to the left-hand side by changing its sign.
\[\Rightarrow \left( x+2 \right)\left( x+4 \right)\]
Therefore, the factors are \[\left( x+2 \right)\left( x+4 \right)\].
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