Question

X³ + 6x -7= 0 là phương trình

Answer 1

Answer:(x-1) *(x^2 + x + 7)

Explanation:

1. rewrite

2. factor the expression

Answer 2

`\huge\boxed{\bf\:x = 1}`

Explanation:

`x ^ { 3 } +6x-7=0`

By using the rational root theorem, all rational roots of a polynomial are in the form p/q, where p divides the constant term -7 & q divides the leading coefficient, 1. Now, let's list out all the possible candidates for p/q by following this theorem.

`(+/-)7, (+/-)1`

Now, let's substitute the value of x as these integers. By using the trial & error method, we can see that..

`x = 1`

This means that 1 factor of the above equation will be:

`x = 1\\\Longrightarrow\:x - 1 = 0`

So, by using the Factor theorem, we know that, x - k will be the factor of the polynomial for each root k. Now, divide x³ + 6x - 7 by x - 1. We'll the value as:

`(x^(3) + 6x - 7)/(x - 1)\\\Longrightarrow \: x^(2)+x+7=0`

We now have a quadratic equation with us. By using the biquadratic formula: -b ± √b² - 4ac / 2a, where,

• a = 1
• b = 1
• c = 7

So,

`x=\frac{-1(+/-)\sqrt{1^(2)-4* 1* 7}}{2} \\x=(-1(+/-)√(-27))/(2)`

Here, we can see that,

`x\in \emptyset`

Then, by listing out the solutions that we found, the value of x will be:

`\boxed{\bf\:x = 1}`

`\rule{150}{2}`