Question

How do you factor the expression $ 5{{x}^{2}}-18x+9 $ ?

Answer 1

Hint: To factor the above expression $ 5{{x}^{2}}-18x+9 $ , first we are going to multiply the constant and coefficient of $ {{x}^{2}} $ . And then find the factors of this multiplication and add and subtract in such a way so that we will get the coefficient of x. Then we will replace 18 by this addition and subtraction of the factors. And then simplify to get the factors.

Complete step by step answer:
The expression is given in the above problem of which we have to find the factors are:
 $ 5{{x}^{2}}-18x+9 $
Now, multiplying the coefficient of $ {{x}^{2}} $ and the constant of the above quadratic expression we get.
 $ 5\times 9 $
Writing factors for the above multiplication we get,
 $ \Rightarrow 5\times 3\times 3\times 1 $
Now, if we take factors 15 and 3 then on the addition of these factors we get 18 so replacing 18 by $ 15+3 $ we get,
 $ \Rightarrow 5{{x}^{2}}-\left( 15+3 \right)x+9 $
Opening the bracket in the above expression we get,
 $ \Rightarrow 5{{x}^{2}}-15x-3x+9 $
Taking 5x as common from the first two terms and -3 from the last two terms we get,
 $ \Rightarrow 5x\left( x-3 \right)-3\left( x-3 \right) $
Now, taking $ x-3 $ as common from the above expression we get,
 $ \Rightarrow \left( x-3 \right)\left( 5x-3 \right) $
Hence, we have factorized the given quadratic expression into $ \left( x-3 \right)\left( 5x-3 \right) $ .

Note:
We can check whether the factors that we are getting are correct or not by multiplying these two factors and see whether we are restoring the quadratic expression or not.
 $ \left( x-3 \right)\left( 5x-3 \right) $
Multiplying the above two factors we get,
 $ \begin{align}
  & x\left( 5x \right)-3x-15x+9 \\
 & \Rightarrow 5{{x}^{2}}-18x+9 \\
\end{align} $
As you can see that we are getting the same quadratic expression which we started with, so the factors which we have calculated are correct.
The other method to check these factors is to put these factors to 0 and calculate the value of x.
 $ \begin{align}
  & x-3=0 \\
 & \Rightarrow x=3 \\
 & 5x-3=0 \\
 & \Rightarrow x=\dfrac{3}{5} \\
\end{align} $
Now, one by one substitute these values of x in the above quadratic expression and check whether on putting these values of x, are you getting the whole expression as 0 or not.
Substituting $ x=3 $ in the above quadratic expression we get,
 $ \begin{align}
  & 5{{\left( 3 \right)}^{2}}-18\left( 3 \right)+9 \\
 & \Rightarrow 5\left( 9 \right)-54+9 \\
 & \Rightarrow 45-54+9 \\
 & \Rightarrow 54-54 \\
 & \Rightarrow 0 \\
\end{align} $
As you can see that $ x=3 $ is satisfying the given quadratic expression so the factor $ x-3 $ which we have found in the above solution is correct. Similarly you can check the other value of x also.