malilevy
Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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

Sagot :

Sueraiuka

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....

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.