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Malilevygo
Answered

Multiply the polynomials: (x – 4)(x2 + 2x – 5)

Question 16 options:

A)

x3 + 6x2+ 3x – 20

B)

x3 + 6x2 + 3x + 20

C)

x3 – 2x2 – 13x – 20

D)

x3 – 2x2 – 13x + 20
Question 17 (5 points)
Subtract: (4x3 + 9xy + 8y) – (3x3 + 5xy – 8y)
Question 17 options:

A)

7x3 + 14xy

B)

x3 + 4xy

C)

7x3 + 14xy + 16y

D)

x3 + 4xy + 16y
Question 18 (5 points)
What are the real solutions to the equation 5x2 + 29x + 20 = 0?
Question 18 options:

A)

x = –5, x = 4∕5

B)

x = –4∕5, x = 5

C)

x = 4∕5, x = 5

Answer :

Sueraiukago

Step-by-step explanation:

Hey there!

Given;

[tex] = (x - 4)( {x}^{2} + 2x - 5)[/tex]

[tex] = x( {x}^{2} + 2x - 5) - 4( {x}^{2} + 2x - 5)[/tex]

[tex] = {x}^{3} + 2 {x}^{2} - 5x - 4 {x}^{2} - 8x + 20[/tex]

[tex] = {x}^{3} - 2 {x}^{2} - 13x + 20[/tex]

Therefore, Option D is correct answer.

Q.no.

Given;

[tex] =( 4 {x}^{3} + 9xy + 8y) - (3 {x}^{3} + 5xy - 8y)[/tex]

[tex] = 4 {x}^{3} + 9xy + 8y - 3 {x}^{3} - 5xy + 8y[/tex]

[tex] = {x}^{3} + 4xy + 16y[/tex]

Therefore, answer is Option D.

Qno.

Given;

[tex]5 {x}^{2} + 29x + 20 = 0[/tex]

[tex]5 {x}^{2} + (25 + 4)x + 20 = 0[/tex]

[tex]5 {x}^{2} + 25x + 4x + 20 = 0[/tex]

[tex]5x(x + 5) + 4(x + 5) = 0[/tex]

(5x + 4)(x + 5) =0

Either (5x+4)= 0,

5x = -4

X = -4/5

Or, X+5 = 0

X = -5.

Therefore, X= -5, -4/5.

Hope it helps....

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