What is the remainder obtained by dividing x7 x5 1 by the generator polynomial x3 1?

Mathematics

2 answers:

posledela3 years ago

4 0

The remainder obtained by dividing on ${x^7} + {x^5} + 1$ the generator polynomial ${x^3} + 1$ is $\boxed{{-x^2} + {x} +1}$

Given:

The generator polynomial is ${x^3} + 1.$

The polynomial is ${x^7} + {x^5} + 1$

Explanation:

There are two methods of dividing the two polynomials.

(1). Synthetic division

(2). Long division method

The fraction can be expressed as,

${\text{Fraction}}= \dfrac{{{x^7} + {x^5} + 1}}{{{x^3} + 1}}$

The denominator of the fraction is ${x^3} + 1$ and the numerator of the fraction is ${x^7} + {x^5} + 1.$

Divide ${x^7} + {x^5} + 1$ by ${x^3} + 1$ to obtain the remainder and the quotient of the polynomial.

The remainder obtained by dividing on ${x^7} + {x^5} + 1$ the generator polynomial ${x^3} + 1$ is $\boxed{{-x^2} + {x} +1}$

Kindly refer the image attached below.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Long division method

Keywords: division, binomial synthetic division, long division method, coefficients, quotients, remainders, numerator, denominator, polynomial, zeros, degree, cubic polynomial, quadratic equation.

vivado [14]3 years ago

3 0

Please, use " ^ " to indicate exponention. x7 and x3 are meaningless.

I must assume you mean x^7 + 0 x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1
and that your "generator polynomial" is x^3 + 0x^2 + 0x + 1. If this is not the case, then you really need to work on your presentation of polynomials.

Divide (x^3 + 0x^2 + 0x + 1 into
x^7 + 0 x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 1

Dividing x^3 into x^7 leaves a partial quotient of x^4. Multiply (x^3 + 1) by this x^4 and write your result under x^7 + 0x^6 + 0x^5 ... + 1
Subtract.

Can you now finish this problem? If not, what do you need to know so that you can finish it?

You might be interested in

Full steps and all reasons, thanks!

Nataly_w [17]

You can focus on the two sub-triangles ABC and ACD

We know that they are both right triangles, because we're given that AC and BD are perpendicular.

The sides AB and AD are the hypothenuses of these triangles, so we know their length through the pythagorean theorem:

$AD^2 = AC^2+CD^2$

$AB^2 = AC^2+BC^2$

But since we're given $BC=CD$ , the expressions for AD and AB are actually the same expression. This proves that AD=AB, and thus that the triangle is isosceles.