Consider the Polynomial P(x)=x^3-5x^2-x+5. is (x-5) a factor?

Mathematics

1 answer:

Alborosie3 years ago

8 0

Answer:

Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

Step-by-step explanation:

Step 1: Factor

p(x) = x^3 - 5x^2 - x + 5

p(x) = (x - 5)(x + 1)(x - 1)

Yes, the remainder is 0 so, x-5 is a factor of p(x)

Answer: Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

You might be interested in

For numbers 19 and 20 you have to solve for the missing angle.

Illusion [34]

Answer:19. 17 degrees

20. 49 degrees

Step-by-step explanation:

Because of the postulate 50+?=67

So ?=17

And 20.

94+?=143

?=49

(Edit)I though you only meant 19 and 20

21. X=8

22. X=12

23.x=8

24.x=7

7 0

3 years ago

Find the pattern and use it to list the nth term in the sequence. 1, 1/32,1/243,1/1024,1/3125

Yuki888 [10]

$1;\ \frac{1}{32};\ \frac{1}{243};\ \frac{1}{1024};\ \frac{1}{3125}\\\\1=\frac{1}{1^5}\\\\\frac{1}{32}=\frac{1}{2^5}\\\\\frac{1}{243}=\frac{1}{3^5}\\\\\frac{1}{1024}=\frac{1}{4^5}\\\\\frac{1}{3125}=\frac{1}{5^5}\\\vdots\\\\a_n=\frac{1}{n^5}\ where\ n\in\mathbb{N^+}$