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3 years ago

What dies it mean to be a factor of a polynomial?

Mathematics

2 answers:

Lapatulllka [165]3 years ago

8 0

Step-by-step explanation:

expresses a polynomial with coefficients in a given field or in the integers as the product of irreducible factors with coefficients in the same domain.

hopes this helps

Gemiola [76]3 years ago

6 0

Answer:

A factor of polynomial P(x) is any polynomial which divides evenly into P(x)

Step-by-step explanation:

For example, x + 2 is a factor of the polynomial x2 – 4. The factorization of a polynomial is its representation as a product its factors.

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kaheart [24]

Answer:

First "order of operations" mistake: step 2

First arithmetic mistake: step 4

Step-by-step explanation:

As we understand Rena's work, she wants to simplify ...

$\left(\dfrac{x^{-3}y^{-2}}{2x^4y^{-4}}\right)^{-3}$

for x = -1 and y = 2.

Her work seems to be ...

Step 1

$\text{Substitute $x=-1$ and $y=2$ into the expression}\\\\\left(\dfrac{(-1)^{-3}2^{-2}}{2(-1)^42^{-4}}\right)^{-3}\qquad\text{no error}$

Step 2

$\text{Simplify the parentheses}\\\\\left(\dfrac{2^4}{2(-1)^4(-1)^32^2}\right)^{-3}=\left(\dfrac{2^2}{2(-1)^7}\right)^{-3}\qquad\text{order of operations error}$

Step 3

$\text{Evaluate the power to a power}\\\\\dfrac{2^{-6}}{2^{-3}(-1)^{21}}\qquad\text{no error}$

Step 4

$\text{Use reciprocals and find the value}\\\\\dfrac{1}{2^32^6(-1)^{21}}=\dfrac{1}{8\cdot 64\cdot (-1)}=\dfrac{-1}{512}\qquad\text{error: $2^3$ is used instead of $2^{-3}$}$

_____

So, the first arithmetic error is in Step 4. However, the order of operations requires exponents be evaluated first. Doing that makes step 2 look like ...

$\left(\dfrac{-\dfrac{1}{4}}{2(1)\dfrac{1}{16}}\right)^{-3}=(-2)^{-3}\qquad\text{proper Step 2}$

We expect your answer is supposed to be Step 4.

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3 years ago

What is the complete factorization of r2 - 6r - 9s2 + 9? A) (r - 3 + 3s)(r - 3 - 3s) B) (r + 3 + 3s)(r - 3 - 3s) C) (r - 3 + 3s)

nikitadnepr [17]

$r^2-6r-9s^2+9=r^2-6r+9-9s^2=r^2-2\cdot r\cdot3+3^2-9s^2\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(r-3)^2-9s^2=(r-3)^2-(3s)^2\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=(r-3-3s)(r-3+3s)\to A$

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