The zeros of a polynomial are the points where the polynomial cross the x-axis.
The factored form of the function is:[tex]\mathbf{(d)\ f(x) =f(x)=(x-3)(x+1)(x-1)(x+2)}[/tex]
The function is given as:
[tex]\mathbf{f(x) = x^4 - x^3 - 7x^2 + x + 6}[/tex]
One of the zeros is at x = 3.
This means that one of the factors is x - 3
Divide f(x) by x - 3
So, we have:
[tex]\mathbf{\frac{f(x)}{x -3} = \frac{x^4 - x^3 - 7x^2 + x + 6}{x-3}}[/tex]
Expand
[tex]\mathbf{\frac{f(x)}{x -3} = \frac{x^4 + 2x^3 -3x^3 - x^2 - 6x^2- 2x +3x +6}{x-3}}[/tex]
Rewrite as:
[tex]\mathbf{\frac{f(x)}{x -3} = \frac{x^4 + 2x^3 - x^2 - 2x -3x^3 - 6x^2 +3x +6}{x-3}}[/tex]
Expand
[tex]\mathbf{\frac{f(x)}{x -3} = \frac{x(x^3 +2x^2 - x - 2)- 3(x^3 +2x^2 - x - 2)}{x-3}}[/tex]
Factor out [tex]\mathbf{(x^3 +2x^2 - x - 2)}[/tex]
[tex]\mathbf{\frac{f(x)}{x -3} = \frac{(x^3 +2x^2 - x - 2)(x - 3)}{x-3}}[/tex]
Multiply both sides by x - 3
[tex]\mathbf{f(x) = (x^3 +2x^2 - x - 2)(x - 3)}[/tex]
Factorize
[tex]\mathbf{f(x) = x(x^2 - 1) + 2(x^2 - 1)(x - 3)}[/tex]
Factor out x^2 - 1
[tex]\mathbf{f(x) = (x + 2) (x^2 - 1)(x - 3)}[/tex]
Express as difference of two squares
[tex]\mathbf{f(x) =(x + 2)(x + 1)(x - 1)(x - 3)}[/tex]
Hence, the factored form of the function is:[tex]\mathbf{(d)\ f(x) =f(x)=(x-3)(x+1)(x-1)(x+2)}[/tex]
Read more about factors of polynomials at:
https://brainly.com/question/12787576
