10 POINTS!!!!!!! write an equation for a line that goes through the point (-3,-4) and has a slope on -2

1 answer:

4 0

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The graph below shows two polynomial functions, f(x) and g(x): Graph of f of x equals x squared minus 2 x plus 1. Graph of g of

IceJOKER [234]

9514 1404 393

Answer:

Step-by-step explanation:

The leading terms of the two functions are ...

f(x): x² (even degree, positive coefficient: 1)

g(x): x³ (odd degree, positive coefficient: 1)

Then it is true that ...

f(x) is an even degree polynomial with a positive leading coefficient

8 0

3 years ago

Factor the polynomial function over the complex numbers. f(x)=x^4-x^3-2x−4 Enter your answer in the box. f(x) = The answer is: (

Vladimir79 [104]

Answer:

Factors: $(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

Step-by-step explanation:

We are given a polynomial:

$f(x)=x^4-x^3-2x-4$

We have to factor the given polynomial into its complex factors.

The factorization can be done as follows:

$f(x)=x^4-x^3-2x-4 = 0\\x^4-x^3-2x-4 = 0\\x^4-4-x^3-2x=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^4-4)-(x^3+2x) = 0\\(x^2+2)(x^2-2)-x(x^2+2) = 0\\(x^2+2)(x^2-2-x) = 0\\(x^2+2)(x^2-x-2) = 0\\(x^2+2)(x^2-2x-+x-2) = 0\\(x^2+2)((x(x-2)+1(x-2))=0\\(x^2+2)(x+1)(x-2)=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^2-(\sqrt{-2})^2)(x+1)(x-2)=0\\(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$