Omar wanted to know if x−5 is a factor of the polynomial p(x)=x3−6x2−x+30. He applied the Factor Theorem and concluded that x−5 is a factor of p(x), as shown in the following work.
Step 1: p(5)=53−6(52)−5+30
Step 2: p(5)=125−150−5+30=0
Step 3: p(5)=0, so the remainder is 0.
Step 4: The remainder is 0, so x−5 is a factor of p(x).
Did Omar make a mistake? If so, in which step does his mistake occur?
Yes, Omar's mistake is in Step 1.
Yes, Omar's mistake is in Step 3.
Yes, Omar's mistake is in Step 2.
Yes, Omar's mistake is in Step 4.
No, Omar did not make any mistakes.
Step 1: p(5)=53−6(52)−5+30 makes no sense
where did those numbers come from?
Step 2: p(5)=125−150−5+30=0
Step 3: p(5)=0, so the remainder is 0.
Step 4: The remainder is 0, so x−5 is a factor of p(x).
those lines were correct.
He means 5^3 and 5^2, not 53 and 52
No, Omar did not make any mistakes.
Omar made a mistake in Step 1. Here's why:
To check if x-5 is a factor of the polynomial p(x) = x^3 - 6x^2 - x + 30, we need to apply the Factor Theorem. According to the theorem, if x-a is a factor of a polynomial, then the polynomial evaluated at a, p(a), should be equal to 0.
In Step 1, Omar substituted x = 5 into the polynomial incorrectly. Instead of evaluating p(5), he evaluated p(5^3) = p(125), p(5^2) = p(25), and p(5), which is incorrect.
To evaluate p(5), we need to substitute x = 5 into p(x):
p(5) = (5^3) - 6(5^2) - 5 + 30
= 125 - 150 - 5 + 30
= 125 - 150 - 5 + 30
= 0 - 5 + 30
= 0 + 25
= 25
Since p(5) is not equal to 0, Omar's conclusion in Step 4 is incorrect. Therefore, the correct answer is:
No, Omar did not make any mistakes.