Multiplying Polynomials

The area of a rectangle is obtained by multiplying its length by its width. Then, let's start by writing the dimensions suggested by Diego's uncle. Length: & 2h-1 Width: & 4h^2-5h+2 If we multiply the binomial by the trinomial, we will get a polynomial modeling the chicken coop area. A(h) = (2h-1)(4h^2-5h+2) To write A(x) in standard form, let's perform the product on the right-hand side.

The product A(h)=(2h-1)(4h^2-5h+2) can also be expanded using the Box Method. Since the first polynomial has two terms and the second has three terms, we will use a table with 2 rows and three columns.

Now, let's fill in the table's cells by multiplying the terms written on the corresponding borders of the table.

Finally, we add all the expressions inside the table and combine like terms.

The volume of a rectangular prism is found by multiplying its three dimensions. Therefore, let's start by writing the dimensions of the given solid. Length: & x+2 Width: & x+1 Height: & x-1 If we multiply the three expressions representing the prism's dimensions, we will get a polynomial representing its volume. V(x) = (x+2)(x+1)(x-1) To write V(x) in standard form, let's multiply the three binomials, starting with the two at the left.

To write V(x) in standard form the three binomials can be multiplied following the next steps.

1. First, multiply the two binomials at the right using the FOIL method.
2. Then, multiply the resulting polynomial by the first binomial.

Let's begin with the first step. Here, we need to multiply the first terms of each binomial. ( x+1)( x-1) ⇓ x* x The empty box is there as a reminder that there are still missing terms. Next, we multiply the outer terms. ( x+1)(x- 1) ⇓ x* x - x* 1 Now, it is the turn of the inner terms. (x+ 1)( x-1) ⇓ x* x - x* 1 + 1* x Then, we multiply the last terms. (x+ 1)(x- 1) ⇓ x* x - x* 1 + 1* x - 1* 1 After performing each of the products and combining like terms, it is obtained the following binomial. (x+1)(x-1) = x^2 - 1 Now, the previous binomial will be multiplied by the binomial (x+2). Again, we can use the FOIL method. ( x+2)( x^2-1) ⇓ x* x^2 We continue multiplying the outer terms. ( x+2)(x^2- 1) ⇓ x* x^2 - x* 1 Next, we multiply the inner terms. (x+ 2)( x^2-1) ⇓ x* x^2 - x* 1 + 2* x^2 Finally, we multiply the last terms. (x+ 2)(x^2- 1) ⇓ x* x^2 - x* 1 + 2* x^2 - 2* 1 Once the products are performed it is obtained the following polynomial. V(x) = x^3 + 2x^2 - x - 2

Although there are some missing terms, we do not need them to find the degree of P(x)Q(x). The reason is that when two polynomials are multiplied, the product is a new polynomial whose degree equals the sum of the degrees of the multiplied polynomials. P(x) &= 3x^4 + 6x^2 + ??? - 4 Q(x) &= 5x^2 - ??? + 2 As we can see, the degree of P(x) is 4 and the degree of Q(x) is 2. With this information, we can find the degree of P(x)Q(x). Degree ofP(x)Q(x) 4+ 2 = 6 Consequently, the degree of the product of the given polynomials is 6.

As in Part A, we do not need to know the missing terms to find the leading coefficient of P(x)Q(x). The reason is that when two polynomials are multiplied, the leading coefficient of the resulting polynomial equals the product of the leading coefficients of the multiplied polynomials. P(x) &= 3x^4 + 6x^2 + ??? - 4 Q(x) &= 5x^2 - ??? + 2 The leading coefficient of P(x) is 3 and the leading coefficient of Q(x) is 5. Let's find their product. Leading Coefficient ofP(x)Q(x) 3* 5 = 15 The leading coefficient of the product of the given polynomials is 15.

In a similar manner, when two polynomials are multiplied, the constant term of the resulting polynomial equals the product of the constant terms of the multiplied polynomials. Then, we only need to know the constant terms of P(x) and Q(x). P(x) &= 3x^4 + 6x^2 + ??? - 4 Q(x) &= 5x^2 - ??? + 2 The constant term of P(x) is -4 and the constant term of Q(x) is 2. Let's find their product. Constant Term ofP(x)Q(x) -4* 2 = -8 The constant term of the product of the given polynomials is -8.