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# Which statements are true about the polynomial function?

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Which statements are true about the polynomial function?

f(x)= x^4+ 5x^3 -x^2 -5x

Statements(Choose all that apply)

f(x) divided by (x+5) has a remainder of 0.

f(5)=0

(x-5) is a factor of f(x)

f(x)=0  when x=-5

Use rational root theorem to determine the factors.

f(x)=2x^3 + x^2 -8x -4

Factors--(x+2), (x+1), (x-2), (x-4), (2x-1), (x+4), (2x+1), (x-1)

My answer-- (x+2) (x+1) (x-2). Am I right? Once again I just need to make sure, thats all, thanks

Guest Nov 9, 2017

#1
+681
+2

1) f(x)= x4+5x3-x2-5x

Statements 1, 3, and 4 are correct. 2 is not because f(5) does not equal 0.

The Rational Root Theorem is:\

Let f(x) be a polynomial with integral coefficients. The only possible rational zeros of f(x) are:

$$\frac{p}{q}$$

where p is a divisor of the constant term and q is a divisor of the leading coefficient.

So, fators of the constant on top and factors of the LC on bottom.

$$\frac{\pm1, \pm2, \pm4}{\pm1,\pm2}$$

Now, pair them together.

Divide all of the top factors by 1 first, then 2.

$$\pm1,\pm2,\pm4,\pm \frac{1}{2}, \pm1,\pm2$$

Repeating factors can be dropped so the possible factors are $$\pm1,\pm2,\pm4,\pm \frac{1}{2}$$ or $$(x+1),(x-1),(x+2),(x-2),(x+4),(x-4),(2x+1),(2x-1)$$.

So your possible factors are correct.

To see which ones are factors, I woud graph it.

This graph shows that $$(-2,0), (-\frac{1}{2},0), (2,0)$$ are the zeros. (Here is the graph: https://www.desmos.com/calculator/fx9umcmzu0)

As factors those are (x+2), (2x+1), and (x-2).

Sort:

#1
+681
+2

1) f(x)= x4+5x3-x2-5x

Statements 1, 3, and 4 are correct. 2 is not because f(5) does not equal 0.

The Rational Root Theorem is:\

Let f(x) be a polynomial with integral coefficients. The only possible rational zeros of f(x) are:

$$\frac{p}{q}$$

where p is a divisor of the constant term and q is a divisor of the leading coefficient.

So, fators of the constant on top and factors of the LC on bottom.

$$\frac{\pm1, \pm2, \pm4}{\pm1,\pm2}$$

Now, pair them together.

Divide all of the top factors by 1 first, then 2.

$$\pm1,\pm2,\pm4,\pm \frac{1}{2}, \pm1,\pm2$$

Repeating factors can be dropped so the possible factors are $$\pm1,\pm2,\pm4,\pm \frac{1}{2}$$ or $$(x+1),(x-1),(x+2),(x-2),(x+4),(x-4),(2x+1),(2x-1)$$.

So your possible factors are correct.

To see which ones are factors, I woud graph it.

This graph shows that $$(-2,0), (-\frac{1}{2},0), (2,0)$$ are the zeros. (Here is the graph: https://www.desmos.com/calculator/fx9umcmzu0)

As factors those are (x+2), (2x+1), and (x-2).

#2
+80935
+1

Very nice,  AT  !!!

CPhill  Nov 9, 2017

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