Question

section 4 – writing a polynomial in standard form
for each problem, given one the zeros and point, write the polynomial out in standard form.
4 points each

1. ( f(x) ) has zeros of ( x = 3 ), ( 4 ) and ( -1 ) and ( f(1) = 24 )

Explanation:

Step1: Form the polynomial with zeros

If a polynomial has zeros \( x = a, x = b, x = c \), then it can be written in factored form as \( f(x)=k(x - a)(x - b)(x - c) \), where \( k \) is a constant. Here, the zeros are \( x = 3, x = 4, x=- 1 \), so the factored form is \( f(x)=k(x - 3)(x - 4)(x + 1) \).

Step2: Find the value of \( k \) using the point \( f(1) = 24 \)

Substitute \( x = 1 \) and \( f(1)=24 \) into the equation:

$$
LATEXBLOCK0
$$

Step3: Expand the polynomial

First, multiply \( (x - 3)(x - 4) \):

$$
LATEXBLOCK1
$$

Then multiply the result by \( (x + 1) \):

$$
LATEXBLOCK2
$$

Now multiply by \( k = 2 \):
\( f(x)=2(x^{3}-6x^{2}+5x + 12)=2x^{3}-12x^{2}+10x + 24 \)

Answer:

\( f(x)=2x^{3}-12x^{2}+10x + 24 \)