How do you factor the expression \[5{{x}^{2}}+8x-21\]?
Answer
612.6k+ 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. 8x, and multiplication is equal to \[5\times 21=5\times 7\times 3=15\times 7\], where \[15-7=8\] 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,
\[5{{x}^{2}}+8x-21\]
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. 8x, and multiplication is equal to \[5\times 21=5\times 7\times 3=15\times 7\], where \[15-7=8\] which is the middle term.
\[\Rightarrow 5{{x}^{2}}+15x-7x+21\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( 5{{x}^{2}}+15x \right)+\left( -7x+21 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow 5x\left( x+3 \right)-7\left( x+3 \right)\]
We can again take the common factor first then the remaining terms, to make a factor, we get
\[\Rightarrow \left( 5x-7 \right)\left( x+3 \right)\]
Therefore, the factors are \[\left( 5x-7 \right)\left( x+3 \right)\].
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,
\[5{{x}^{2}}+8x-21\]
Where, a = 5, b = 8, c = -21.
We can substitute the above values in quadratic formula.
\[\begin{align}
& \Rightarrow x=\dfrac{-8\pm \sqrt{64+420}}{10} \\
& \Rightarrow x=\dfrac{-8\pm \sqrt{484}}{10} \\
& \Rightarrow x=\dfrac{-8\pm 22}{10} \\
& \Rightarrow x=\dfrac{7}{5},-3 \\
\end{align}\]
Where,\[x=-3,x=\dfrac{7}{5}\]
We can take the term in the right-hand side, to the left-hand side by changing its sign.
\[\left( 5x-7 \right)\left( x+3 \right)\]
Therefore, the factors are \[\left( 5x-7 \right)\left( x+3 \right)\].
Complete step by step solution:
We know that the given expression is,
\[5{{x}^{2}}+8x-21\]
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. 8x, and multiplication is equal to \[5\times 21=5\times 7\times 3=15\times 7\], where \[15-7=8\] which is the middle term.
\[\Rightarrow 5{{x}^{2}}+15x-7x+21\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( 5{{x}^{2}}+15x \right)+\left( -7x+21 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow 5x\left( x+3 \right)-7\left( x+3 \right)\]
We can again take the common factor first then the remaining terms, to make a factor, we get
\[\Rightarrow \left( 5x-7 \right)\left( x+3 \right)\]
Therefore, the factors are \[\left( 5x-7 \right)\left( x+3 \right)\].
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,
\[5{{x}^{2}}+8x-21\]
Where, a = 5, b = 8, c = -21.
We can substitute the above values in quadratic formula.
\[\begin{align}
& \Rightarrow x=\dfrac{-8\pm \sqrt{64+420}}{10} \\
& \Rightarrow x=\dfrac{-8\pm \sqrt{484}}{10} \\
& \Rightarrow x=\dfrac{-8\pm 22}{10} \\
& \Rightarrow x=\dfrac{7}{5},-3 \\
\end{align}\]
Where,\[x=-3,x=\dfrac{7}{5}\]
We can take the term in the right-hand side, to the left-hand side by changing its sign.
\[\left( 5x-7 \right)\left( x+3 \right)\]
Therefore, the factors are \[\left( 5x-7 \right)\left( x+3 \right)\].
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