Chapter 1.5, Problem 3E

(a)

To determine

To fill: The blank in the statement “The solutions of the equation $x^2\left(x-4\right)=0$ are ____”.

Answer to Problem 3E

The complete statement is “The solutions of the equation $x^2\left(x-4\right)=0$ are $\underset{\_}{x=0\text{or }x=4}$”.

Explanation of Solution

Property used:

Zero-product property:

If $ab=0$, then $a=0\text{or}b=0$, or both.

Calculation:

Use zero product property, to calculate the solutions of the equation $x^2\left(x-4\right)=0$.

$\begin{array}{l}{x}^2\left(x-4\right)=0\\ x^2=0\text{or}\left(x-4\right)=0\\ x=0\text{and}x=4\end{array}$

Thus, the complete statement is “The solutions of the equation $x^2\left(x-4\right)=0$ are $\underset{\_}{x=0\text{or }x=4}$”.

(b)

To determine

To fill: The blank in the statement “To solve the equation $x^3-4x^2=0$, we _____ the left-hand side”.

Answer to Problem 3E

The complete statement is “To solve the equation $x^3-4x^2=0$, we $\underset{\_}{\text{factor}}$ the left-hand side”.

Explanation of Solution

Note that, the expression on the left side of a quadratic equation in the standard form $a{x}^2+bx+c=0,a\ne 0$ can be factored into the product of two first-degree polynomials.

Then, by using the zero-product property set each factor equal to zero. Then, the resulting equations are solved to obtain the given quadratic equation.

Thus, the given equation can be solved by factoring the expression $x^3-4x^2$ as follows.

$\begin{array}{l}{x}^3-4x^2=0\\ x^2\left(x-4\right)=0\left[\text{Take}{x}^2\text{as}\text{common factor}\right]\\ x^2=0\text{or}x-4=0\left[\text{Zero}\text{product}\text{property}\right]\\ x=0\text{or}x=4\end{array}$

Therefore, the complete statement is “To solve the equation $x^3-4x^2=0$, we $\underset{\_}{\text{factor}}$ the left-hand side”.