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The length of a rectangle is 20 units more than its width. The area of the rectangle is x4 - 100.Which statement about the width of the rectangle is true?x2 – 10 because the area expression can be rewritten as (x2 – 10)( x2 + 10) which equals (x2 – 10)((x2 – 10) + 20).x2 + 10 because the area expression can be rewritten as (x2 + 10)( x2 – 10) which equals (x2 + 10)((x2 + 10) – 20).x2 – 30 because the area expression can be rewritten as (x2 + 10)( x2 – 10) which equals (x2 + 10)((x2 – 30) + 20).x2 + 30 because the area expression can be rewritten as (x2 – 10)( x2 + 10) which equals (x2 – 10)((x2 + 30) – 20).

Question

The length of a rectangle is 20 units more than its width. The area of the rectangle is x4 - 100.Which statement about the width of the rectangle is true?x2 – 10 because the area expression can be rewritten as (x2 – 10)( x2 + 10) which equals (x2 – 10)((x2 – 10) + 20).x2 + 10 because the area expression can be rewritten as (x2 + 10)( x2 – 10) which equals (x2 + 10)((x2 + 10) – 20).x2 – 30 because the area expression can be rewritten as (x2 + 10)( x2 – 10) which equals (x2 + 10)((x2 – 30) + 20).x2 + 30 because the area expression can be rewritten as (x2 – 10)( x2 + 10) which equals (x2 – 10)((x2 + 30) – 20).

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Solution

To solve the problem, let's break it down step by step:

  1. Identify the given information:

    • The length of the rectangle is 20 units more than its width.
    • The area of the rectangle is given by the expression x4100 x^4 - 100 .
  2. Express the length and width in terms of a variable:

    • Let the width of the rectangle be w w .
    • Then, the length of the rectangle is w+20 w + 20 .
  3. Set up the equation for the area:

    • The area of the rectangle is given by the product of its length and width.
    • Therefore, w×(w+20)=x4100 w \times (w + 20) = x^4 - 100 .
  4. Factor the given area expression:

    • The expression x4100 x^4 - 100 can be factored using the difference of squares: x4100=(x210)(x2+10) x^4 - 100 = (x^2 - 10)(x^2 + 10)
  5. Match the factored form to the product of width and length:

    • We need to match w×(w+20) w \times (w + 20) to (x210)(x2+10) (x^2 - 10)(x^2 + 10) .
  6. Compare the factors:

    • If we let w=x210 w = x^2 - 10 , then the length w+20 w + 20 becomes: (x210)+20=x2+10 (x^2 - 10) + 20 = x^2 + 10
    • This matches the factored form (x210)(x2+10) (x^2 - 10)(x^2 + 10) .

Therefore, the correct statement about the width of the rectangle is:

  • x210 x^2 - 10 because the area expression can be rewritten as (x210)(x2+10) (x^2 - 10)(x^2 + 10) which equals (x210)((x210)+20) (x^2 - 10)((x^2 - 10) + 20) .

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