If f(x)=x^3+9x^2+2x-48 and f(-3)=0 . th algebraically.

To find the other zeros of the polynomial \( f(x) = x^3 + 9x^2 + 2x - 48 \) given that \( f(-3) = 0 \), we can use synthetic division to factor out \( (x + 3) \) from the polynomial.<br /><br />### Step 1: Synthetic Division<br /><br />We will perform synthetic division using \( x = -3 \).<br /><br />1. Write down the coefficients of \( f(x) \):<br /> \[<br /> 1, 9, 2, -48<br /> \]<br /><br />2. Set up the synthetic division:<br /> \[<br /> \begin{array}{r|rrrr}<br /> -3 & 1 & 9 & 2 & -48 \\<br /> & & -3 & -18 & 48 \\<br /> \hline<br /> & 1 & 6 & -16 & 0 \\<br /> \end{array}<br /> \]<br /><br />Here's the step-by-step process:<br />- Bring down the leading coefficient (1).<br />- Multiply -3 by 1 and write it under the next coefficient: \( -3 \times 1 = -3 \).<br />- Add the second coefficient (9) and -3: \( 9 + (-3) = 6 \).<br />- Multiply -3 by 6 and write it under the next coefficient: \( -3 \times 6 = -18 \).<br />- Add the third coefficient (2) and -18: \( 2 + (-18) = -16 \).<br />- Multiply -3 by -16 and write it under the next coefficient: \( -3 \times -16 = 48 \).<br />- Add the fourth coefficient (-48) and 48: \( -48 + 48 = 0 \).<br /><br />The bottom row gives us the coefficients of the quotient polynomial:<br />\[<br />x^2 + 6x - 16<br />\]<br /><br />### Step 2: Solve the Quotient Polynomial<br /><br />Now we need to find the zeros of the quadratic polynomial \( x^2 + 6x - 16 \).<br /><br />We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = -16 \).<br /><br />\[<br />x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1}<br />\]<br />\[<br />x = \frac{-6 \pm \sqrt{36 + 64}}{2}<br />\]<br />\[<br />x = \frac{-6 \pm \sqrt{100}}{2}<br />\]<br />\[<br />x = \frac{-6 \pm 10}{2}<br />\]<br /><br />This gives us two solutions:<br />\[<br />x = \frac{-6 + 10}{2} = \frac{4}{2} = 2<br />\]<br />\[<br />x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8<br />\]<br /><br />### Step 3: List All Zeros<br /><br />The zeros of the polynomial \( f(x) = x^3 + 9x^2 + 2x - 48 \) are:<br />\[<br />x = -3, x = 2, x = -8<br />\]