Answer:
[tex]\huge\boxed{\bf\:x = 1}[/tex]
Step-by-step explanation:
[tex]x ^ { 3 } +6x-7=0[/tex]
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.
[tex](+/-)7, (+/-)1[/tex]
Now, let's substitute the value of x as these integers. By using the trial & error method, we can see that..
[tex]x = 1[/tex]
This means that 1 factor of the above equation will be:
[tex]x = 1\\\Longrightarrow\:x - 1 = 0[/tex]
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:
[tex]\frac{x^{3} + 6x - 7}{x - 1}\\\Longrightarrow \: x^{2}+x+7=0[/tex]
We now have a quadratic equation with us. By using the biquadratic formula: -b ± √b² - 4ac / 2a, where,
So,
[tex]x=\frac{-1(+/-)\sqrt{1^{2}-4\times 1\times 7}}{2} \\x=\frac{-1(+/-)\sqrt{-27}}{2}[/tex]
Here, we can see that,
[tex]x\in \emptyset[/tex]
Then, by listing out the solutions that we found, the value of x will be:
[tex]\boxed{\bf\:x = 1}[/tex]
[tex]\rule{150}{2}[/tex]
Answer:(x-1) *(x^2 + x + 7)
Step-by-step explanation:
1. rewrite
2. factor the expression
