Chapter 8, Problem 28PT

To determine

To factor the given polynomial

Answer to Problem 28PT

  $4(4x+5)(4x+5)$

Explanation of Solution

Given:

Polynomial: $16x^2+40x+25$

Formula used:

Roots of the polynomial of the form $a{x}^2+bx+c=a(x-x_1)(x-x_2)$ are

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the given polynomial,

  $\Rightarrow 16x^2+40x+25$

Applying the above formula for $a=16,b=40,c=25$

  $\begin{array}{l}\Rightarrow x=\frac{-40\pm \sqrt{{40}^2-4(16)(25)}}{2(16)}\\ \Rightarrow x=\frac{-40\pm \sqrt{1600-1600}}{32}\\ \Rightarrow x=\frac{-40}{32}\\ \Rightarrow x=\frac{-5}{4}\end{array}$

Therefore, the roots are

  $\Rightarrow x_1=x_2=\frac{-5}{4}$

Therefore, the polynomial can be factorised as,

  $\begin{array}{l}\Rightarrow (x-(\frac{-5}{4}))(x-(\frac{-5}{4}))\\ \Rightarrow (x+\frac{5}{4})(x+\frac{5}{4})\\ \Rightarrow 4(4x+5)(4x+5)\end{array}$

Conclusion:

Therefore, the polynomial is factored as

  $4(4x+5)(4x+5).$