Note: This is not middle school math. If it is, god your screwed.
Answer:
A) x-2
Step-by-step explanation:
To find a factor of an equation, you must divide the equation by a binomial (which is the answer choices). Also, when you divide, it is only declared factorable when there is NO REMAINDER. Not even if it's 1 remainder.
ANSWER
A) x-2
EXPLANATION
Using the Remainder Theorem, if x=a is a factor of p(x), then p(a)=0.
Let
[tex]p(x) = {x}^{4} + 3 {x}^{3} - 5 {x}^{2} + 2x - 24[/tex]
[tex]p(2) = {2}^{4} + 3 {(2)}^{3} - 5 {(2)}^{2} + 2(2) - 24[/tex]
[tex]p(2) = 16+ 24 - 20 + 4- 24[/tex]
[tex]p(2) = 24- 24 = 0[/tex]
Since
[tex]p(2) =0[/tex]
x-2 is a factor of
[tex]p(x) = {x}^{4} + 3 {x}^{3} - 5 {x}^{2} + 2x - 24[/tex]
For the other options,
[tex]p(3) =99[/tex]
[tex]p(5) =861[/tex]
Hence x-3 and x-5 are not factors of the given function.
The correct answer is A.
