Question

1. show how you can find the other factor.

(chart with x, 1 rows and x³, x², 3x, 2 columns)

1. can you factor any further? if so, write the completely factored polynomial below.

(handwritten: (x+1)(x+1)(x+2))

1. solve g(x) = 0. what do these solutions tell you about the graph of g?
2. use the area model to divide g(x) = x³ + 4x² + 5x + 2 by x + 3.

(chart with 3 rows and 4 columns)

1. is x + 3 a factor of the polynomial? how can you tell?

Explanation:

Question 5: Solve \( g(x) = 0 \). What do these solutions tell you about the graph of \( g \)?

Assuming from Question 4 that \( g(x) = (x + 1)^2(x + 2) \) (the factored form), we solve \( g(x) = 0 \):

Step 1: Set the factored polynomial to zero

We have \( (x + 1)^2(x + 2) = 0 \). By the Zero - Product Property, if a product of factors is zero, then at least one of the factors must be zero. So we set each factor equal to zero:

• For \( (x + 1)^2=0 \), taking the square root of both sides (or using the property that if \( a^2 = 0 \), then \( a = 0 \)), we get \( x+1 = 0\), so \( x=-1 \). Since the exponent of \( (x + 1) \) is 2, the root \( x=-1 \) has a multiplicity of 2.
• For \( (x + 2)=0 \), we solve for \( x \) and get \( x=-2 \).

Step 2: Interpret the solutions for the graph

The solutions to \( g(x)=0 \) are the x - intercepts of the graph of \( y = g(x) \). A root with multiplicity 2 (like \( x=-1 \)) means that the graph touches the x - axis at \( x=-1 \) and turns around (since the multiplicity is even). A root with multiplicity 1 (like \( x=-2 \)) means that the graph crosses the x - axis at \( x=-2 \).

Step 1: Recall the Remainder Theorem

For a polynomial \( g(x) \) and a linear divisor \( x - a \), the remainder when \( g(x) \) is divided by \( x - a \) is \( g(a) \). If \( x + 3=x-(-3) \) is a factor of \( g(x) \), then the remainder when \( g(x) \) is divided by \( x + 3 \) should be zero (by the Factor Theorem, which is a special case of the Remainder Theorem: \( x - a \) is a factor of \( g(x) \) if and only if \( g(a)=0 \)).

Step 2: Substitute \( x=-3 \) into \( g(x)=x^{3}+4x^{2}+5x + 2 \)

Calculate \( g(-3)\):

$$
LATEXBLOCK0
$$

Since \( g(-3)=-4
eq0 \), by the Factor Theorem, \( x + 3 \) is not a factor of \( g(x) \). We can also check the result of the division in Question 6. If we divide \( g(x)=x^{3}+4x^{2}+5x + 2 \) by \( x + 3 \) using the area model, we will find that there is a non - zero remainder (in this case, the remainder is - 4), which also shows that \( x + 3 \) is not a factor.

Answer:

The solutions to \( g(x)=0 \) are \( x=-1 \) (with multiplicity 2) and \( x=-2 \). These solutions are the x - intercepts of the graph of \( g \). The graph touches the x - axis at \( x = - 1 \) (because of the even multiplicity) and crosses the x - axis at \( x=-2 \) (because of the odd multiplicity).

Question 7: Is \( x + 3 \) a factor of the polynomial? How can you tell?

We can use the Remainder Theorem. The Remainder Theorem states that if we divide a polynomial \( g(x) \) by \( x - a \), the remainder is \( g(a) \). If we are dividing by \( x+3=x-(-3) \), then \( a=-3 \). We can also use the result of the division in Question 6 (using the area model) or direct substitution.