
How do you factor $ {{x}^{2}}+14x+45 $ ?
Answer
496.2k+ views
Hint: In the question, we need to factorize the given equation. We can observe that the given equation is a quadratic equation. Generally, we will follow different methods for different types of equations. For the quadratic equation which is in the form of $ a{{x}^{2}}+bx+c $ , we will split the middle term into two parts like $ b={{x}_{1}}+{{x}_{2}} $ such that $ {{x}_{1}}.{{x}_{2}}=ac $ . So, we will first calculate the value of $ ac $ and we will write the factors of the calculated $ ac $ and choose two factors such that they satisfy the above-mentioned conditions. After getting the values of $ {{x}_{1}} $ , $ {{x}_{2}} $ we will split the middle term and simplify the equation by taking appropriate terms as common. After taking the terms as common we will get our required result.
Complete step by step answer:
Given that, $ {{x}^{2}}+14x+45 $ .
Comparing the above equation with $ a{{x}^{2}}+bx+c $ , then we can write
Coefficient of $ {{x}^{2}} $ is $ a=1 $ .
Coefficient of $ x $ is $ b=14 $ .
Constant is $ c=45 $ .
Now the value of $ ac $ is given by
$ \begin{align}
& ac=1\left( 45 \right) \\
& \Rightarrow ac=45 \\
\end{align} $
Here we have the value of $ ac $ as $ 45 $ . We know that the factors for $ 45 $ are $ 1 $ , $ 3 $ , $ 5 $ , $ 9 $ , $ 15 $ , $ 45 $ . From the above factors we can observe that
$ \begin{align}
& 5+9=14 \\
& 5\times 9=45 \\
\end{align} $
So, we can split the middle term $ 14x $ as $ 14x=5x+9x $ . Substituting this value in the given equation, then we will get
$ {{x}^{2}}+14x+45={{x}^{2}}+5x+9x+45 $
Taking $ x $ as common from $ {{x}^{2}}+5x $ and $ 9 $ as common from $ 9x+45 $ , then the above equation is modified as
$ {{x}^{2}}+14x+45=x\left( x+5 \right)+9\left( x+5 \right) $
Again, taking $ x+5 $ as common on RHS of the above equation, then we will get
$ \Rightarrow {{x}^{2}}+14x+45=\left( x+5 \right)\left( x+9 \right) $
Hence the factors of the given equation are $ x+5 $ and $ x+9 $ .
Note:
In this problem, they have only asked to factorize the given equation, so we have ended our procedure after calculating the factors. If they have asked to solve the equation or asked to find the roots, then we need to equate the given equation to zero and we will find the values of $ x $.
Complete step by step answer:
Given that, $ {{x}^{2}}+14x+45 $ .
Comparing the above equation with $ a{{x}^{2}}+bx+c $ , then we can write
Coefficient of $ {{x}^{2}} $ is $ a=1 $ .
Coefficient of $ x $ is $ b=14 $ .
Constant is $ c=45 $ .
Now the value of $ ac $ is given by
$ \begin{align}
& ac=1\left( 45 \right) \\
& \Rightarrow ac=45 \\
\end{align} $
Here we have the value of $ ac $ as $ 45 $ . We know that the factors for $ 45 $ are $ 1 $ , $ 3 $ , $ 5 $ , $ 9 $ , $ 15 $ , $ 45 $ . From the above factors we can observe that
$ \begin{align}
& 5+9=14 \\
& 5\times 9=45 \\
\end{align} $
So, we can split the middle term $ 14x $ as $ 14x=5x+9x $ . Substituting this value in the given equation, then we will get
$ {{x}^{2}}+14x+45={{x}^{2}}+5x+9x+45 $
Taking $ x $ as common from $ {{x}^{2}}+5x $ and $ 9 $ as common from $ 9x+45 $ , then the above equation is modified as
$ {{x}^{2}}+14x+45=x\left( x+5 \right)+9\left( x+5 \right) $
Again, taking $ x+5 $ as common on RHS of the above equation, then we will get
$ \Rightarrow {{x}^{2}}+14x+45=\left( x+5 \right)\left( x+9 \right) $
Hence the factors of the given equation are $ x+5 $ and $ x+9 $ .
Note:
In this problem, they have only asked to factorize the given equation, so we have ended our procedure after calculating the factors. If they have asked to solve the equation or asked to find the roots, then we need to equate the given equation to zero and we will find the values of $ x $.
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