Question

I need help as soon as possible. Find the real or imaginary solutions of each equation by factoring. ( I need an explanation I don’t understand)

27x^3+1=0

And

X^3-27=0

Answer 1

Explanation:

These questions are both to practice factoring a "sum/difference of two cubes" Someone else back in the 1500's figured out how to factor this. We just have to follow the pattern. Memorize this pattern.

The sum or difference of two cubes factors into a binomial and a trinomial. For the binomial just peel off the exponent (the 3rd power) from both terms and thats the first factor. For the trinomial, change the power to a 2 for both terms, but put in between them another term that is the the two terms multiplied together.

Lets look at your #6)

27x^3 + 1

Rewrite 27 as 3^3 and 1 as 1^3.

(3x)^3 + 1^3

FIRST FACTOR:

Just rip off the exponents:

(3x + 1)

SECOND FACTOR:

(3x)^2 - 3x + 1^2

^^^see the first and last terms are the 3x and the 1 but now they have a exponent, power of 2.

In between the multiply the 3x and the 1 times together.

The binomial has the Same sign as your original question. The trinomial has signs Opposite, then the last sign is Alwayspositive. The acronym SOAP can help you remember. The + or - in the factors.

I wish this was just a factoring problem. But we have to solve too.

3x+1=0

3x = -1

x = -1/3

AND

(3x)^2 + 3x + 1 =0

9x^2 + 3x + 1 = 0

oh no, we can't factor this, we have to use Quadratic formula: a=9, b=3, c=1

x= (-3+-root(3^2-4*9*1))/2*9

x = (-3 +- root(-27))/18

x= (-3 +- 3iroot3)/18

x = (-1 +- i*root3)/6

This is two solutions. Bc of the +- in the final answer. Some texts/programs/teacher might require you to write an imaginary number in the form a+bi

So it would be:

-1/6 + (i*root3)/6 and

-1/6 - (i*root3)/6

And from above the third solution is -1/3

Answer 2

Let's solve these questions :

`27x^3+1=0`

Taking 1 to the right side as 0 has no use in the equation,

`↪ 27{x}^(3) = - 1`

`↪ {x}^(3) = - (1)/(27)`

`↪x = \sqrt[3]{ - (1)/(27) }`

Now, as 27 is the cube of 3,

`↪x = - (1)/(3)`

So,
`x = (1)/(3)`

Now Second question,

`x^3-27=0`

Taking the negative 27 to the right side, it becomes positive.

`↪ {x}^(3) = 27`

`↪x = \sqrt[3]{27}`

`↪x = 3`

So value of x is 3

~ Benjemin360