Consider the polynomial functions defined by f(x)= (x-1)(x-2)(x+3), g(x)=x^3+2 x^2-x-2, and

Consider the polynomial functions defined by f(x)=...

Consider the polynomial functions defined by $f(x)=$ $(x-1)(x-2)(x+3), \quad g(x)=x^{3}+2 x^{2}-x-2, \quad$ and $h(x)=3 x^{3}+6 x^{2}-3 x-6$ (a) What is the value of $f(1) ?$ (b) For what values, other than $1,$ is $f(x)=0 ?$ (c) Verify that $g(-1)=g(1)=g(-2)=0$ (d) Based on your answer from part (c), what do you think is the factored form of $g(x) ?$ Verify your answer by multiplying these factors and comparing with $g(x)$ (e) Using your answer from part (d), what is the factored formof $h(x) ?$ (f) Based on what you have learned in this exercise, fill in the blank: If $f$ is a polynomial and $f(a)=0$ for some number $a$ then one factor of the polynomial is

Consider the polynomial functions defined by $f(x)=$ $(x-1)(x-2)(x+3), \quad g(x)=x^{3}+2 x^{2}-x-2, \quad$ and $h(x)=3 x^{3}+6 x^{2}-3 x-6$
(a) What is the value of $f(1) ?$
(b) For what values, other than $1,$ is $f(x)=0 ?$
(c) Verify that $g(-1)=g(1)=g(-2)=0$
(d) Based on your answer from part (c), what do you think is the factored form of $g(x) ?$ Verify your answer by multiplying these factors and comparing with $g(x)$
(e) Using your answer from part (d), what is the factored formof $h(x) ?$
(f) Based on what you have learned in this exercise, fill in the blank: If $f$ is a polynomial and $f(a)=0$ for some number $a$ then one factor of the polynomial is

Key Concepts

Factoring Polynomials

Factoring involves writing a polynomial as a product of simpler polynomials or linear factors. This process is facilitated by first identifying the zeros of the polynomial and then using the Factor Theorem to express the polynomial as a product of factors corresponding to these zeros.

Factor Theorem

The Factor Theorem asserts that if a polynomial f(x) equals zero for x = a, then (x - a) is a factor of f(x). It is a central tool for polynomial factorization, particularly in finding the factors once the zeros of the polynomial have been identified.

Verification Through Polynomial Multiplication

After factoring a polynomial, it is important to verify the factorization by multiplying the factors back together. This step ensures that the expanded form is equivalent to the original polynomial, confirming the accuracy of the factorization process.

Polynomial Evaluation

This concept involves substituting a specific value for the variable in a polynomial and performing the arithmetic to calculate the result. It is fundamental to understanding how polynomials behave at particular points and forms the basis of checking solutions or understanding the graph of the polynomial.

Zeros and the Zero Product Property

A polynomial has zeros (or roots) where its value is equal to zero. The zero product property states that if a product of factors equals zero, then at least one of the factors must be zero. This principle is used to determine the factors of a polynomial once its zeros are known.

Transcript

00:01 Alright, so for the first part of this problem, let's evaluate f of 1 from this function over here.

00:09 When we put x equals 1, we get 0 multiplied by 1 minus 2 is negative 1, multiply by 1 plus 3 is 4.

00:23 Now 4 times negative 1 is negative 4, but anything multiply by 0 just evaluates to 0.

00:30 So we can say that f of 1 equals 0.

00:39 Now what other values of x would f the function f of x evaluate to 0? we can see that the function would evaluate to 0 if any of these terms is 0.

00:54 So when x equal to 2, the middle term would be 0, and the function would evaluate to 0.

01:02 So f of 2 is also 0.

01:07 And then when x equals negative 3, the third term over here would evaluate to a 0.

01:14 So f of negative 3 would also be equal to 0.

01:20 Moving on to the next part.

01:23 Let's evaluate g of x at those different points that we've been given.

01:29 Just rewrite g of x over here, x cubed, plus 2x.

01:37 Square minus x minus 2.

01:42 So at negative 1, we have g of negative 1 equals negative 1 cubed is a negative 1 plus 2 times negative 1 whole square is just a 1 minus negative 1 minus 2.

02:06 So we have negative 1 plus 2 plus 1 negative 2.

02:15 So the 2's cancel out and the 1's cancel out and we're left with 0.

02:20 So that checks out.

02:22 Next, let's see what happens at x equals 1.

02:25 We have 1 cubed is 1 plus 2 times 1 minus 1 minus 2.

02:33 Which is just 1 plus 2 minus 1 minus 2.

02:37 And again, everything nicely cancels out, and we get g of 1 is 0...