Question

X³ + 6x -7= 0 là phương trình

Answer 1

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

Step-by-step explanation:

1. rewrite

2. factor the expression

Answer 2

Answer:

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

Step-by-step 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:

$\frac{x^{3} + 6x - 7}{x - 1}\\\Longrightarrow \: x^{2}+x+7=0$

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,

$x=\frac{-1(+/-)\sqrt{1^{2}-4\times 1\times 7}}{2} \\x=\frac{-1(+/-)\sqrt{-27}}{2}$

Here, we can see that,

$x\in \emptyset$

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

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

$\rule{150}{2}$