Matemáticas, pregunta formulada por B0R, hace 8 meses

x6+10x3+25 respuesta

Respuestas a la pregunta

Contestado por Nickhomsom
3

Explicación paso a paso:

x6-10x3+25

Final result :

(x3 - 5)2

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x3" was replaced by "x^3". 1 more similar replacement(s).

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((x6) - (2•5x3)) + 25

Step 2 :

Trying to factor by splitting the middle term

2.1 Factoring x6-10x3+25

The first term is, x6 its coefficient is 1 .

The middle term is, -10x3 its coefficient is -10 .

The last term, "the constant", is +25

Step-1 : Multiply the coefficient of the first term by the constant 1 • 25 = 25

Step-2 : Find two factors of 25 whose sum equals the coefficient of the middle term, which is -10 .

-25 + -1 = -26

-5 + -5 = -10 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -5 and -5

x6 - 5x3 - 5x3 - 25

Step-4 : Add up the first 2 terms, pulling out like factors :

x3 • (x3-5)

Add up the last 2 terms, pulling out common factors :

5 • (x3-5)

Step-5 : Add up the four terms of step 4 :

(x3-5) • (x3-5)

Which is the desired factorization

Trying to factor as a Difference of Cubes:

2.2 Factoring: x3-5

Theory : A difference of two perfect cubes, a3 - b3 can be factored into

(a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =

a3+a2b+ab2-ba2-b2a-b3 =

a3+(a2b-ba2)+(ab2-b2a)-b3 =

a3+0+0+b3 =

a3+b3

Check : 5 is not a cube !!

Ruling : Binomial can not be factored as the difference of two perfect cubes

Polynomial Roots Calculator :

2.3 Find roots (zeroes) of : F(x) = x3-5

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 1 and the Trailing Constant is -5.

The factor(s) are:

of the Leading Coefficient : 1

of the Trailing Constant : 1 ,5

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 -6.00

-5 1 -5.00 -130.00

1 1 1.00 -4.00

5 1 5.00 120.00

Polynomial Roots Calculator found no rational roots

Trying to factor as a Difference of Cubes:

2.4 Factoring: x3-5

Check : 5 is not a cube !!

Ruling : Binomial can not be factored as the difference of two perfect cubes

Polynomial Roots Calculator :

2.5 Find roots (zeroes) of : F(x) = x3-5

See theory in step 2.3

In this case, the Leading Coefficient is 1 and the Trailing Constant is -5.

The factor(s) are:

of the Leading Coefficient : 1

of the Trailing Constant : 1 ,5

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 -6.00

-5 1 -5.00 -130.00

1 1 1.00 -4.00

5 1 5.00 120.00

Polynomial Roots Calculator found no rational roots

Multiplying Exponential Expressions :

2.6 Multiply (x3-5) by (x3-5)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x3-5) and the exponents are :

1 , as (x3-5) is the same number as (x3-5)1

and 1 , as (x3-5) is the same number as (x3-5)1

The product is therefore, (x3-5)(1+1) = (x3-5)2

Final result :

(x3 - 5)2

Contestado por lalisaMLily
2

Explicación paso a paso:

x⁶+10x³+25

x³. 5

2(x³+5)

(x³+5)²

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