Chapter 6.7, Problem 2CYU

To determine

Solve the given equation. State the number and type of roots.

Answer to Problem 2CYU

The equation $x^3+12x^2+32x=0$ has three roots, all three are real 0, -8, -4.

Explanation of Solution

Given information:

The equation $x^3+12x^2+32x=0$ .

Formula used:

Splitting the middle term: To solve a quadratic equation $a{x}^2+bx+c=0$ , the middle term of equation is splitted in such a way that the product of coefficient of its parts is equal to product of coefficient of first and third term.

  $x^3+12x^2+32x=0$

Taking x common in LHS

  $\begin{array}{l}{x}^3+12x^2+32x=0\\ \\ x\left(x^2+12x+32\right)=0\end{array}$

Recall that in splitting the middle term, the middle term of equation is splitted in such a way that the product of coefficient of its parts is equal to product of coefficient of first and third term.

  $\begin{array}{l}x\left(x^2+12x+32\right)=0\\ \\ x\left(x^2+8x+4x+32\right)=0\\ \\ x\left(x\left(x+8\right)+4\left(x+8\right)\right)=0\\ \\ x\left(x+4\right)\left(x+8\right)=0\end{array}$

By the above equation, we get that one, two or all the three terms $\left(x+4\right)$ , $x$ and $\left(x+8\right)$ are zero,

Putting $\left(x+4\right)$ as zero,

  $\begin{array}{l}x+4=0\\ \\ x=-4\end{array}$

Putting $\left(x+8\right)$ as zero,

  $\begin{array}{l}x+8=0\\ \\ x=-8\end{array}$

Putting $x$ as zero,

  $x=0$

The equation $x^3+12x^2+32x=0$ has three roots, all three are real 0, -8, -4.