ms. wilson draws a model of the factorization of a polynomial with integer factors. her model is partially complete. which equation is represented by ms. wilson’s model? n2 3n 40 = (n – 8)(n – 5) n2 13n 40 = (n 8)(n 5) n2 40n 13 = (n 8)(n 5) n2 40n 3 = (n – 8)(n – 5)

I'll just factor the above equation.

x² + 18x + 80

x² ⇒ x * x
80
can be:
1 x 80
2 x 40
4 x 20
5 x 16
8 x 10 Correct pair

(x+8)(x+10)
x(x+10) +8(x+10) ⇒ x² + 10x + 8x + 80 = x² + 18x + 80

x+8 = 0
x = -8

x+10 = 0
x = -10

x = -8

(-8)² + 18(-8) + 80 = 0
64 - 144 + 80 = 0
144 - 144 = 0
0 = 0

(-10)² + 18(-10) + 80 = 0
100 - 180 + 80 = 0
180 - 180 = 0
0 = 0

I think the algebra tiles will not be a good tool to use to factor the quadratic equation because the equation is not a perfect square quadratic equation.

calculista calculista · Answer 2 · 2018-08-05T00:56:10+02:00

the correct question in the attached figure

We proceed to analyze each case

case a) [tex] n^{2}+ 3n + 40 = (n- 8)(n- 5) [/tex]

Expand the right side and compare with the left side

so

[tex] (n- 8)(n- 5)=n^{2} -5n-8n+40\\ (n- 8)(n- 5)=n^{2} -13n+40
[/tex]

[tex] n^{2}+ 3n + 40 [/tex] is not equal to [tex] n^{2} -13n+40 [/tex]

therefore

case a) is not the equation represented by ms. wilson’s model

case b) [tex] n^{2} + 13n + 40 = (n + 8)(n + 5) [/tex]

Expand the right side and compare with the left side

so

[tex] (n+ 8)(n+ 5)=n^{2} +5n+8n+40\\ (n+ 8)(n+ 5)=n^{2} +13n+40
[/tex]

[tex] n^{2} + 13n + 40 [/tex] is equal to [tex] n^{2} + 13n + 40 [/tex]

therefore

case b) is the equation represented by ms. wilson’s model

case c) [tex] n^{2} + 40n + 13 = (n + 8)(n + 5) [/tex]

Expand the right side and compare with the left side

so

[tex] (n+ 8)(n+ 5)=n^{2} +5n+8n+40\\ (n+ 8)(n+ 5)=n^{2} +13n+40
[/tex]

[tex] n^{2} + 40n + 13 [/tex] is not equal to [tex] n^{2} + 13n + 40 [/tex]

therefore

case c) is not the equation represented by ms. wilson’s model

case d) [tex] n^{2} + 40n + 3 = (n - 8)(n - 5) [/tex]

Expand the right side and compare with the left side

so

[tex] (n- 8)(n- 5)=n^{2} -5n-8n+40\\ (n- 8)(n- 5)=n^{2} -13n+40
[/tex]

[tex] n^{2} + 40n + 3 [/tex] is not equal to [tex] n^{2} - 13n + 40 [/tex]

therefore

case d) is not the equation represented by ms. wilson’s model

therefore

the answer is the case b) [tex] n^{2} + 13n + 40 = (n + 8)(n + 5) [/tex]