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kondaur [170]
4 years ago
15

How many real roots would f(x)= x^3+2x^2+2x-5 have

Mathematics
1 answer:
goldfiish [28.3K]4 years ago
7 0

Answer:

One real root

Step-by-step explanation:

By the fundamental theorem of algebra, an nth degree polynomial has n-possible real roots.

If there are complex roots, then the the complex roots come in pairs.

Therefore the number of possible real roots of f(x)=x^3+2x^2+2x-5 are 3 real roots with no complex pairs or 1 real root with a complex pair.

By Descartes rule of signs, there is only one change of sign in the polynomial (+ to -).

Hence there is only one positive real root.

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

There is change is sign two times but we can not have even number of real roots for this polynomial

Therefore there is only real root.

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A whole number that rounds to 500 if we are rounding to the nearest hundred
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4 years ago
When 5(2x+3)-(x+4) simplified is?
emmasim [6.3K]

Answer:

9x + 11

Step-by-step explanation:

5(2x+3)-(x+4)

10x+15-x-4        Use distributive property

10x-x +15-4       Combine like terms

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Hope this helps!

3 0
3 years ago
Solve the equation below. Show the steps that lead to your final solution.
Stolb23 [73]

Answer:

this is how u would do it

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8 0
3 years ago
If f(x)= 9x-8 and g(x)=x^3,what is (g o f)(1)?
lesya692 [45]

(f∘g)(3) = f(g(3))


Find g(3) by substituting x=3 in g(x)


g(x)=x²-1

g(3)=3²-1

g(3)=9-1

g(3)=8


f*g(3) = f(g(3)) = f(8)


find f(8) by substituting x=8 in f(x):


f(x)=2x+3

f(8)=2(8)+3

f(8)=16+3

f(8)=19


So f∘g(3) = 19


Sorry I am wrong



4 0
4 years ago
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