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Karo-lina-s [1.5K]
3 years ago
7

What are the coefficients for the expression

2%20" id="TexFormula1" title="4x4 + x3 - 2x2 - 6x + 2 " alt="4x4 + x3 - 2x2 - 6x + 2 " align="absmiddle" class="latex-formula">


​
Mathematics
2 answers:
In-s [12.5K]3 years ago
8 0

6x - 2 + 2x = -2 + 4x + 8

Simplifying

6x + -2 + 2x = -2 + 4x + 8

Reorder the terms:

-2 + 6x + 2x = -2 + 4x + 8

Combine like terms: 6x + 2x = 8x

-2 + 8x = -2 + 4x + 8

Reorder the terms:

-2 + 8x = -2 + 8 + 4x

Combine like terms: -2 + 8 = 6

-2 + 8x = 6 + 4x

Solving

-2 + 8x = 6 + 4x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-4x' to each side of the equation.

-2 + 8x + -4x = 6 + 4x + -4x

Combine like terms: 8x + -4x = 4x

-2 + 4x = 6 + 4x + -4x

Combine like terms: 4x + -4x = 0

-2 + 4x = 6 + 0

-2 + 4x = 6

Add '2' to each side of the equation.

-2 + 2 + 4x = 6 + 2

Combine like terms: -2 + 2 = 0

0 + 4x = 6 + 2

4x = 6 + 2

Combine like terms: 6 + 2 = 8

4x = 8

Divide each side by '4'.

x = 2

Simplifying

x = 2

kirill115 [55]3 years ago
5 0
4x^4+x3-2x^2-6x+2
step by step:

(1): "x2" was replaced by "x^2". 2 more similar replacement(s).

2. ((((4•(x4))+(x3))-2x2)-6x)+2 (equation at the end of step 1)

3. equation at the end of step 2:

(((22x4 + x3) - 2x2) - 6x) + 2

4. polynomial roots caclulator:
3.1 Find roots (zeroes) of : F(x) = 4x4+x3-2x2-6x+2
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 4 and the Trailing Constant is 2.

final result:

4x4 + x3 - 2x2 - 6x + 2
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VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

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b) The recurrence for the coefficients c_k is

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so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

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