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appendix b: extra practice multiplying binomials and factoring trinomia…

Question

appendix b: extra practice
multiplying binomials and factoring trinomials

find the product of the binomials.

  1. $(x - 9)(x + 9) = \underline{\quad\quad}$
  2. $(9x + 2)(3x + 4) = \underline{\quad\quad}$
  3. $(5x - 1)(x - 6) = \underline{\quad\quad}$
  4. $(y + 7)^2 = \underline{\quad\quad}$
  5. $(2x - 8)^2 = \underline{\quad\quad}$
  6. $(6x^2 - 8x)(-x + 9) = \underline{\quad\quad}$
  7. $(4x + 10)(2x - 5) = \underline{\quad\quad}$
  8. $(3y + 6)(9y - 4) = \underline{\quad\quad}$
  9. $(2x + 2y)(3x + 4) = \underline{\quad\quad}$
  1. complete the matrix. then write the equation with the binomials and their product.
$\boldsymbol{\cdot}$$\boldsymbol{x}$$\boldsymbol{-5}$
$\boldsymbol{12x}$
  1. if the length of a playground is $(x - 4)$ feet and the area is $(x^2 - 8x + 16)$ square feet, can the width of the playground be $(x + 4)$ feet? if not, what is the width? explain.

Explanation:

Response

Let's solve question 13 step by step:

Step 1: Recall the formula for the area of a rectangle

The area \( A \) of a rectangle is given by the product of its length \( l \) and width \( w \), i.e., \( A = l \times w \).

Step 2: Substitute the given values

We know that the length \( l = (x - 4) \) feet and the area \( A = (x^2 - 8x + 16) \) square feet. We need to check if the width \( w = (x + 4) \) feet satisfies the area formula.

First, let's factor the area expression \( x^2 - 8x + 16 \). Notice that this is a perfect square trinomial:
\[
x^2 - 8x + 16 = (x - 4)^2
\]
( because \( (a - b)^2 = a^2 - 2ab + b^2 \), here \( a = x \), \( b = 4 \), so \( -2ab = -8x \) and \( b^2 = 16 \))

Step 3: Check the width

If the length is \( (x - 4) \) and the area is \( (x - 4)^2 \), then the width \( w \) should be:
\[
w=\frac{A}{l}=\frac{(x - 4)^2}{(x - 4)}
\]
Assuming \( x
eq4 \) (since length can't be zero), we can simplify this by canceling one \( (x - 4) \) term:
\[
w = x - 4
\]

Now, let's check if \( (x + 4) \) can be the width. Multiply the length \( (x - 4) \) by the proposed width \( (x + 4) \):
\[
(x - 4)(x + 4)=x^2 - 16
\]
But the area is \( x^2 - 8x + 16 \), which is not equal to \( x^2 - 16 \) (unless \( -8x + 16=-16 \), which would mean \( -8x=-32 \) or \( x = 4 \), but at \( x = 4 \), the length is zero, which doesn't make sense for a playground). So \( (x + 4) \) cannot be the width.

Final Answer

The width cannot be \( (x + 4) \) feet. The width is \( (x - 4) \) feet because the area \( x^2 - 8x + 16 \) factors to \( (x - 4)^2 \), and when we divide the area by the length \( (x - 4) \) (for \( x
eq4 \)), we get the width \( (x - 4) \). Multiplying the length \( (x - 4) \) by \( (x + 4) \) gives \( x^2 - 16 \), which is not equal to the given area \( x^2 - 8x + 16 \).

Answer:

Let's solve question 13 step by step:

Step 1: Recall the formula for the area of a rectangle

The area \( A \) of a rectangle is given by the product of its length \( l \) and width \( w \), i.e., \( A = l \times w \).

Step 2: Substitute the given values

We know that the length \( l = (x - 4) \) feet and the area \( A = (x^2 - 8x + 16) \) square feet. We need to check if the width \( w = (x + 4) \) feet satisfies the area formula.

First, let's factor the area expression \( x^2 - 8x + 16 \). Notice that this is a perfect square trinomial:
\[
x^2 - 8x + 16 = (x - 4)^2
\]
( because \( (a - b)^2 = a^2 - 2ab + b^2 \), here \( a = x \), \( b = 4 \), so \( -2ab = -8x \) and \( b^2 = 16 \))

Step 3: Check the width

If the length is \( (x - 4) \) and the area is \( (x - 4)^2 \), then the width \( w \) should be:
\[
w=\frac{A}{l}=\frac{(x - 4)^2}{(x - 4)}
\]
Assuming \( x
eq4 \) (since length can't be zero), we can simplify this by canceling one \( (x - 4) \) term:
\[
w = x - 4
\]

Now, let's check if \( (x + 4) \) can be the width. Multiply the length \( (x - 4) \) by the proposed width \( (x + 4) \):
\[
(x - 4)(x + 4)=x^2 - 16
\]
But the area is \( x^2 - 8x + 16 \), which is not equal to \( x^2 - 16 \) (unless \( -8x + 16=-16 \), which would mean \( -8x=-32 \) or \( x = 4 \), but at \( x = 4 \), the length is zero, which doesn't make sense for a playground). So \( (x + 4) \) cannot be the width.

Final Answer

The width cannot be \( (x + 4) \) feet. The width is \( (x - 4) \) feet because the area \( x^2 - 8x + 16 \) factors to \( (x - 4)^2 \), and when we divide the area by the length \( (x - 4) \) (for \( x
eq4 \)), we get the width \( (x - 4) \). Multiplying the length \( (x - 4) \) by \( (x + 4) \) gives \( x^2 - 16 \), which is not equal to the given area \( x^2 - 8x + 16 \).