Chapter 6.6, Problem 30PPS

To determine

To calculate: the factors for polynomial function $f(x)=20x^3-47x^2+8x+12$

Answer to Problem 30PPS

  $(x+0.4)(x-0.7)(x-2)$ .

Explanation of Solution

Given information:

  

Formula used:

Factor Theorem: If $f(x)$ is a polynomial of degree $n\ge 1$ and a is any real number , then $(x-a)$ is a factor of $f(x)$ , if $f(a)=0$ .

The graph of the polynomial function of degree n has at most $n-1$ turning points.

Calculation:

According to the graph and the function ,

The degree of function is 3. Therefore, there are two turning points as observed in the graph. The graph has three x -intercepts. The graph has one y -intercept at $y=12$ .

At point $x=-0.4$ , the value of function is zero. This implies $(x+0.4)$ is a factor of polynomial function.

At point $x=0.7$ , the value of function is zero. This implies $(x-0.7)$ is a factor of polynomial function.

At point $x=2$ , the value of function is zero. This implies $(x-2)$ is a factor of polynomial function.

Further, since the degree is 3 there will be 3 factors of the polynomial.

Therefore we have , $f(x)=(x+0.4)(x-0.7)(x-2)$

Hence, the factors are $(x+0.4)(x-0.7)(x-2)$ .