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2 years ago

-11 X +4+ 8X -4+ 3X equals

Mathematics

2 answers:

maksim [4K]2 years ago

6 0

Answer:

Step-by-step explanation:

First what you want to do is add together the common numbers

So, take all of the X's and add them together

(-11, 8, and 3)

-11x + 8x + 3x

-3x + 3x

So that all evens out

Now just find the other values

(4 and -4)

4 - 4

This equation equals 0

Hope this helps!!!

-Unicorns110504

*Please mark brainliest*

tatyana61 [14]2 years ago

6 0

Answer:

$0$

Step-by-step explanation:

Combine all like-terms to arrive at your answer:

$0 = 0 ± 0$

I am joyous to assist you anytime.

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The following equation is being multiplied by the LCD. Complete the multiplication to eliminate the denominators

frosja888 [35]

Answer:

$(x+2)(x-2) -1(3x)=(x-3)(x-2)$

Step-by-step explanation:

x+2/3x - 1/x-2 = x-3/3x

$\frac{x+2}{3x}-\frac{1}{x-2}=\frac{x-3}{3x}$

LCD is (3x)(x-2). so we multiply the whole equation by LCD

$(3x)(x-2)(\frac{x+2}{3x}-\frac{1}{x-2}=\frac{x-3}{3x})$

We multiply each term by LCD

$(3x)(x-2)(\frac{x+2}{3x})-(3x)(x-2)(\frac{1}{x-2})=(3x)(x-2)(\frac{x-3}{3x})$

$(x+2)(x-2) -1(3x)=(x-3)(x-2)$

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2 years ago

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Are the leap years divisible by 2

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Any year that is evenly divisible by 4 is a leap year.

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Ayuda necesito resolver este problema con procedimiento ;)

Paraphin [41]

$x^3-2x^2+x-1$ is one of the prime factors of the polynomial

<h3>How to factor the expression?</h3>

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

$x^8 - 3x^6 + x^4 - 2x^3 - 1$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2 -3x^2 +4x^2-2x+2x-1$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)$

Factor out x^6-2x^5+2x^3-3x^2+2x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)$

Express x^2 + 2x + 1 as a perfect square

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)$

Expand the polynomial by adding 0's in the form of +a - a

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)$

Rearrange the terms

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)$

Factorize the expression

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))$

Factor out x^3-2x^2+x-1

$x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)$

One of the factors of the above polynomial is $x^3-2x^2+x-1$ .

This is the same as the option (c)

Hence, $x^3-2x^2+x-1$ is one of the prime factors of the polynomial