Question

How do you factor the trinomial $1-14x+49{{x}^{2}}$?

Answer 1

Hint: For this problem we need to calculate the factors of the given equation. We can observe that the given equation is a quadratic equation which is not in standard form. So, we will rearrange the terms in the given equation to achieve the standard form. After getting standard form $a{{x}^{2}}+bx+c$, we will write the values of $a$, $b$, $c$ by comparing both the equations. Now we will calculate the value of $ac$ and split the middle term as $b={{x}_{1}}+{{x}_{2}}$ where ${{x}_{1}}$, ${{x}_{2}}$ are the factors of the value $ac$. After that we will take appropriate terms as common and simplify the equation to get the result.

Complete step by step solution:
Given equation, $1-14x+49{{x}^{2}}$.
Rearranging the terms in the above equation to achieve standard form of the equation, then we will get
$\Rightarrow 1-14x+49{{x}^{2}}=49{{x}^{2}}-14x+1$
Comparing the above equation with standard form $a{{x}^{2}}+bx+c$, then we will have the values
$a=49$, $b=-14$, $c=1$.
Now the value of $ac$ will be $\Rightarrow ac=1\left( 49 \right)=49$ and factors of the $49$ are $1$, $7$, $49$.
From the above factors we can write
$\begin{align}
  & -7-7=-14 \\
 & -7\times -7=49 \\
\end{align}$
So, we are going to split the middle term $-14x$ as $-7x-7x$, then the given equation is modified as
$\Rightarrow 1-14x+49{{x}^{2}}=49{{x}^{2}}-7x-7x+1$
Taking $7x$ as common from the first two terms and taking $-1$ as common from last two terms, then we will get
$\Rightarrow 1-14x+49{{x}^{2}}=7x\left( 7x-1 \right)-1\left( 7x-1 \right)$
Taking $7x-1$ as common in the above equation, then we will get
$\Rightarrow 1-14x+49{{x}^{2}}=\left( 7x-1 \right)\left( 7x-1 \right)$
Hence the factors of the given equation $1-14x+49{{x}^{2}}$ are $7x-1$, $7x-1$. The typical graph of the given equation will be

Note: For this problem we can also use another method which uses the algebraic formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. So, we can simplify the given equation as follows
$\begin{align}
  & \Rightarrow 1-14x+49{{x}^{2}}={{\left( 7x \right)}^{2}}-2\left( 7x \right)\left( 1 \right)+{{1}^{2}} \\
 & \Rightarrow 1-14x+49{{x}^{2}}={{\left( 7x-1 \right)}^{2}} \\
\end{align}$
From this method also we have the factors of the given equation as $7x-1$, $7x-1$.