find all real-number solutions
x^4-7x^3-8x-56=0
2007-07-10 06:51:49 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
isn't it +8x ?
2007-07-10 06:58:52 · answer #1 · answered by FifiLone 2 · 0⤊ 1⤋
I think you may have made an error copying your problem. Did you mean x^4-7x^3-8x+56?
If so, we can use factoring by grouping:
look at the first two terms x^4-7x^3, look for a common term and factor. This gives us x^3(x-7) now look at the last two terms -8x+56, if we factor out -8 we get x-7, therefore we have a common factor of x-7. So x^4-7x^3-8x+56=x^3(x-7) + (-8)(x-7) and by reverse distribution equals (x^3-8)(x-7).
Hope that helps.
2007-07-10 14:17:26 · answer #2 · answered by kmac90604 1 · 0⤊ 0⤋
x^4-7x^3-8x-56=0
The equation has two real roots at approximately x = -1.696 and x = 7.295.
Too bad that the -56 wasn't a plus 56 because then we could have factored the polynomial.
2007-07-10 14:23:08 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
x^4 - 7x^3 - 8x - 56
There are no rational solutions, so finding roots is pretty difficult. I would just use a calculator, which gave me roots:
x = -1.69611 or x = 7.29462
2007-07-10 13:59:48 · answer #4 · answered by pki15 4 · 0⤊ 0⤋
I thought it was summer......hmmm
2007-07-10 13:59:14 · answer #5 · answered by Smartypants 3 · 0⤊ 0⤋