Math, asked by jitroy2001, 10 months ago

m^5-13m^3+26m^2+82m+104​=0

Answers

Answered by Rocky1951
1
m5-13m3+26m2+82m+104=0

One solution was found :

m = -4
Step by step solution :

Step 1 :

Equation at the end of step 1 :

((((m5)-(13•(m3)))+(2•13m2))+82m)+104 = 0

Step 2 :

Equation at the end of step 2 :

((((m5) - 13m3) + (2•13m2)) + 82m) + 104 = 0

Step 3 :

m5-13m3+26m2+82m+104
can be divided with m+4

Polynomial Long Division :

3.2 Polynomial Long Division
Dividing : m5-13m3+26m2+82m+104

Quotient : m4-4m3+3m2+14m+26 Remainder: 0

Equation at the end of step 3 :

(m4 - 4m3 + 3m2 + 14m + 26) • (m + 4) = 0

Step 4 :

Theory - Roots of a product :

4.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

Quartic Equations :

4.2 Solve m4-4m3+3m2+14m+26 = 0

In search of an interavl at which the above polynomial changes sign, from negative to positive or the other wayaround.

Method of search: Calculate polynomial values for all integer points between m=-20 and m=+20

No interval at which a change of sign occures has been found. Consequently, Bisection Approximation can not be used. As this is a polynomial of an even degree it may not even have any real (as opposed to imaginary) roots

Solving a Single Variable Equation :

4.3 Solve : m+4 = 0

Subtract 4 from both sides of the equation :
m = -4

One solution was found :

m = -4

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.
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