Question

question if f(x)=4x^3 - 28x^2 + 21x + 18 and f(6)=0, then find all of the zeros of f(x) algebraically.

Explanation:

Step1: Use factor - theorem

Since \(f(6)=0\), \((x - 6)\) is a factor of \(f(x)=4x^{3}-28x^{2}+21x + 18\). Use polynomial long - division or synthetic division. Let's use synthetic division. The coefficients of \(f(x)\) are \(4,-28,21,18\).

6	4	-28	21	18
	4	-4	-3	0

The quotient is \(4x^{2}-4x - 3\).

Step2: Factor the quadratic

We factor the quadratic \(4x^{2}-4x - 3\). We need to find two numbers that multiply to \(4\times(-3)=-12\) and add up to \(-4\). The numbers are \(-6\) and \(2\).
\[4x^{2}-4x - 3=4x^{2}-6x+2x - 3 = 2x(2x - 3)+1(2x - 3)=(2x - 3)(2x+1)\]

Step3: Find the zeros

Set \(f(x)=(x - 6)(2x - 3)(2x + 1)=0\).
If \(x-6 = 0\), then \(x = 6\); if \(2x-3=0\), then \(x=\frac{3}{2}\); if \(2x + 1=0\), then \(x=-\frac{1}{2}\).

Answer:

\(x = 6,x=\frac{3}{2},x=-\frac{1}{2}\)