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Crazy boy [7]
4 years ago
5

What are all of the real roots of the following polynomial? f(x) = x5 + 5x4 - 5x3 - 25x2 + 4x + 20

Mathematics
2 answers:
Nadusha1986 [10]4 years ago
8 0
F(x)=x^5 + 5*x^4 - 5*x^3 - 25*x^2 + 4*x + 20

By examining the coefficients of the polynomial, we find that
1+5-5-25+4+20=0 => (x-1) is a factor
Now, reverse the sign of coefficients of odd powers,
-1+5+5-25-4+20=0 => (x+1) is a factor

By the rational roots theorem, we can continue to try x=2, or factor x-2=0
2^5+5(2^4)-5(2^3)-25(2^2)+4(2)+20=0
and similarly f(-2)=0
So we have found four of the 5 real roots.
The remainder can be found by synthetic division as x=-5


Answer: The real roots of the given polynomial are: {-5,-2,-1.1.2}

IgorC [24]4 years ago
3 0

Answer:

The real roots are the following=

Step-by-step explanation:

{ -5,-2 , -1.1.2 }

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zmey [24]

Answer:

the answer would be 36.

Step-by-step explanation:

and here go you would get it

R(-5) = -3*(-5) - 4

R(-5) = 15 - 4

R(-5) = 11

Q(X)=3X+3

Q(11) = 3*11 + 3

Q(11) = 33 + 3

Q(11) = 36

So, (Q(R(-2) = 36

hope it is correct..

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Answer:

<h2>(2a − 5 + b) · 5</h2><h2>10×(a − 2.5 + 0.5b)</h2><h2>(−2a + 5 − b) ⋅ (−5)</h2>

Step-by-step explanation:

10a-25+5b\\\\-2(-5a-25+5b)=(-2)(-5a)+(-2)(-25)+(-2)(5b)\\=10a+50-10b\qquad NOT\\\\(2a-5+b)(5)=(2a)(5)+(-5)(5)+(b)(5)=10a-25+5b\qquad YES\\\\10(a-2.5+0.5b)=(10)(a)+(10)(-2.5)+(10)(0.5b)\\=10a-25+5b\qquad YES\\\\(-2a+5-b)(-5)=(-2a)(-5)+(5)(-5)+(-b)(-5)\\=10a-25+5b\qquad YES\\\\-10(-a-2.5-0.5b)=(-10)(-a)+(-10)(-2.5)+(-10)(-0.5b)\\=10a+25+5b\qquad NOT

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