Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
[tex]\text{Let } f(x) = x^3 + 2x^2 + kx - 6[/tex]
As (x + 1) is a factor, it follows that
[tex]f(-1) = 0[/tex]
[tex]\implies (-1)^3 + 2(-1)^2 + k(-1) - 6 = 0[/tex]
[tex]-1 + 2 -k - 6 = 0[/tex]
[tex]-k - 5 = 0[/tex]
[tex]k = -5[/tex]
[tex]\implies f(x) = x^3 + 2x^2 - 5x - 6[/tex]
Now we can use long division to find what is left of f(x) after it is divided by (x + 1). (Apologies, this is the best way I can represent long division on Brainly at this current time - I hope it's clear)
x^2 + x - 6
x + 1 ( x^3 + 2x^2 - 5x - 6
x^3 + x^2
x^2 - 5x
x^2 + x
-6x - 6
-6x - 6
0
So the remainder when f(x) is divided by (x + 1) is
[tex]x^2 + x - 6[/tex]
Factorising this we get
[tex](x + 3)(x - 2)[/tex]
So the three factors of f(x) are (x + 1), (x + 3) and (x - 2).
As (x + 1) is a factor, it follows that
[tex]f(-1) = 0[/tex]
[tex]\implies (-1)^3 + 2(-1)^2 + k(-1) - 6 = 0[/tex]
[tex]-1 + 2 -k - 6 = 0[/tex]
[tex]-k - 5 = 0[/tex]
[tex]k = -5[/tex]
[tex]\implies f(x) = x^3 + 2x^2 - 5x - 6[/tex]
Now we can use long division to find what is left of f(x) after it is divided by (x + 1). (Apologies, this is the best way I can represent long division on Brainly at this current time - I hope it's clear)
x^2 + x - 6
x + 1 ( x^3 + 2x^2 - 5x - 6
x^3 + x^2
x^2 - 5x
x^2 + x
-6x - 6
-6x - 6
0
So the remainder when f(x) is divided by (x + 1) is
[tex]x^2 + x - 6[/tex]
Factorising this we get
[tex](x + 3)(x - 2)[/tex]
So the three factors of f(x) are (x + 1), (x + 3) and (x - 2).
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.