Evaluate: f(x)=x^3+7x^2+7x-15

Expression: $f\left( x \right)={x}^{3}+7{x}^{2}+7x-15$

To find the $x$-intercept/zero, substitute $f\left( x \right)=0$

$0={x}^{3}+7{x}^{2}+7x-15$

Move the expression to the left-hand side and change its sign

$0-{x}^{3}-7{x}^{2}-7x+15=0$

Removing $0$ doesn't change the value, so remove it from the expression

$-{x}^{3}-7{x}^{2}-7x+15=0$

Write $-7{x}^{2}$ as a difference

$-{x}^{3}+{x}^{2}-8{x}^{2}-7x+15=0$

Write $-7x$ as a difference

$-{x}^{3}+{x}^{2}-8{x}^{2}+8x-15x+15=0$

Factor out $-{x}^{2}$ from the expression

$-{x}^{2} \times \left( x-1 \right)-8{x}^{2}+8x-15x+15=0$

Factor out $-8x$ from the expression

$-{x}^{2} \times \left( x-1 \right)-8x \times \left( x-1 \right)-15x+15=0$

Factor out $-15$ from the expression

$-{x}^{2} \times \left( x-1 \right)-8x \times \left( x-1 \right)-15\left( x-1 \right)=0$

Factor out $-\left( x-1 \right)$ from the expression

$-\left( x-1 \right) \times \left( {x}^{2}+8x+15 \right)=0$

Write $8x$ as a sum

$-\left( x-1 \right) \times \left( {x}^{2}+5x+3x+15 \right)=0$

Factor out $x$ from the expression

$-\left( x-1 \right) \times \left( x \times \left( x+5 \right)+3x+15 \right)=0$

Factor out $3$ from the expression

$-\left( x-1 \right) \times \left( x \times \left( x+5 \right)+3\left( x+5 \right) \right)=0$

Factor out $x+5$ from the expression

$-\left( x-1 \right) \times \left( x+5 \right) \times \left( x+3 \right)=0$

Change the signs on both sides of the equation

$\left( x-1 \right) \times \left( x+5 \right) \times \left( x+3 \right)=0$

When the product of factors equals $0$, at least one factor is $0$

$\begin{array} { l }x-1=0,\\x+5=0,\\x+3=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=1,\\x+5=0,\\x+3=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=1,\\x=-5,\\x+3=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=1,\\x=-5,\\x=-3\end{array}$

The equation has $3$ solutions

$\begin{array} { l }x_1=-5,& x_2=-3,& x_3=1\end{array}$

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