Question

How do you factor the trinomial $ 2{{x}^{2}}+10x+25? $

Answer 1

Hint: To find the factor of given trinomial which is basically a quadratic polynomial we will first take 2 as common and make coefficient of $ {{x}^{2}} $ as 1 and then we will find the root of the given quadratic polynomial obtained by using the formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ , let us say that root is $ \alpha ,\beta $ then the factor of the given polynomial obtained after taking common 2 will be $ \left( x-\alpha \right)\left( x-\beta \right) $ .

Complete step by step answer:
We will start solving the given polynomial by first recalling the Euclidean division of polynomials.
We know that when we divide a divided polynomial $ p\left( x \right) $ with degree $ n $ by some divisor polynomial $ d\left( x \right) $ with degree $ m\le n $ then we get the quotient polynomial $ q\left( x \right) $ of degree $ n-m $ and the remainder polynomial as $ r\left( x \right) $ .We use Euclidean division formula and can write as
 $ p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right) $
We also know that if the reminder polynomial is zero then we call $ d\left( x \right),q\left( x \right) $ factor polynomial of $ p\left( x \right) $ or simply factors of $ p\left( x \right) $ . If $ {{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right).......{{p}_{k}}\left( x \right) $ are $ k $ factors of $ p\left( x \right) $ then we say $ {{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right).......{{p}_{k}}\left( x \right) $ is factorized completely if none of the factors $ {{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right).......{{p}_{k}}\left( x \right) $ can be factorized further.
So, to find the factor of the given trinomial $ 2{{x}^{2}}+10x+25 $ , we will at first take 2 as common factor and then we can write $ 2{{x}^{2}}+10x+25 $ as:
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}+10x+25 \\
 & \Rightarrow 2\left( {{x}^{2}}+5x+\dfrac{25}{2} \right) \\
\end{align}\]
Now, to further factorize will first find the root of \[\left( {{x}^{2}}+5x+\dfrac{25}{2} \right)=0\], and if the root of the equation is $ \alpha ,\beta $ , then the factor of the given polynomial obtained after taking common 2 will be $ \left( x-\alpha \right)\left( x-\beta \right) $ .
Now, for the general quadratic equation $ a{{x}^{2}}+bx+c=0 $ , we know that that its roots are given by:
 $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
So, for the quadratic equation \[\left( {{x}^{2}}+5x+\dfrac{25}{2} \right)=0\], roots are equal to:
 $ \Rightarrow x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 1\times \dfrac{25}{2}}}{2\times 1} $
 $ \Rightarrow x=\dfrac{-5\pm \sqrt{{{5}^{2}}-2\times 1\times 25}}{2\times 1} $
 $ \Rightarrow x=\dfrac{-5\pm \sqrt{25-50}}{2\times 1}=\dfrac{-5\pm \sqrt{-25}}{2} $
We can see that roots are imaginary as we are getting the term $ \sqrt{-1} $ which we write as iota $ i $ .
 $ \Rightarrow x=\dfrac{-5\pm \sqrt{-1}\times \sqrt{25}}{2} $
 $ \Rightarrow x=\dfrac{-5\pm 5i}{2} $
 $ \therefore x=\dfrac{-5+5i}{2},\dfrac{-5-5i}{2} $
Hence, $ \left( x-\dfrac{\left( -5+5i \right)}{2} \right),\left( x-\dfrac{\left( -5-5i \right)}{2} \right) $ are the factor of the polynomial \[\left( {{x}^{2}}+5x+\dfrac{25}{2} \right)\].
So, we can write $ 2{{x}^{2}}+10x+25=2\left( x-\dfrac{\left( -5+5i \right)}{2} \right)\left( x-\dfrac{\left( -5-5i \right)}{2} \right) $
Hence, $ 2,\left( x-\dfrac{\left( -5+5i \right)}{2} \right),\left( x-\dfrac{\left( -5-5i \right)}{2} \right) $ are the factors of the trinomial $ 2{{x}^{2}}+10x+25. $
This is our required solution.

Note:
We note that the highest power on the variable is called a degree of the polynomial. If degree is 1 we call the polynomial a linear polynomial. If the polynomial is single term then we call the polynomial a monomial and if it has two terms it is called binomial. And if the factor has three terms then it is called trinomial. Hence, $ 2{{x}^{2}}+10x+25 $ is a trinomial. Also, note that when we equate factor with zero then we will get the root or we can say the solution of that polynomial.