Answer:
The non real solutions are [tex]\displaystyle \left(\frac{2}{3} +\frac{1}{3} \sqrt{5} i\right)\ and\ \left(\frac{2}{3} -\frac{1}{3} \sqrt{5} i\right)[/tex].
Step-by-step explanation:
We can find the non real solutions for [tex]3x^5+25x^4+26x^3-82x^2+76x=48[/tex] by factorizing the polynomial.
Based on the graph (see the 1st attached picture), the real solutions are {-6, -4, 1}. Therefore some of the factors of the polynomial are:
Hence we can write the polynomial as:
[tex]3x^5+25x^4+26x^3-82x^2+76x-48=0[/tex]
[tex](x+6)(x+4)(x-1)(ax^2+bx+c)=0[/tex]
To find the [tex](ax^2+bx+c)[/tex], we can use the polynomial division either by long division or synthetic division:
[tex](x+6)(x+4)(x-1)(ax^2+bx+c)=3x^5+25x^4+26x^3-82x^2+76x-48[/tex]
[tex](x^3+9x^2+14x-24)(ax^2+bx+c)=3x^5+25x^4+26x^3-82x^2+76x-48[/tex]
[tex](ax^2+bx+c)=(3x^5+25x^4+26x^3-82x^2+76x-48)\div(x^3+9x^2+14x-24)[/tex]
[tex](ax^2+bx+c)=(3x^2-2x+2)[/tex] → see the 2nd attached picture
Now, to find the non real solutions, we can use the abc formula for the factor [tex](3x^2-2x+2)[/tex]:
[tex]\boxed{x=\frac{-b\pm\sqrt{b^2-4ac} }{2a} }[/tex]
[tex]\displaystyle x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(2)} }{2(3)}[/tex]
[tex]\displaystyle x=\frac{4\pm\sqrt{-20} }{6}[/tex]
[tex]\displaystyle x=\frac{4\pm2\sqrt{5} i}{6}[/tex]
[tex]\bf\displaystyle x=\left(\frac{2}{3} +\frac{1}{3} \sqrt{5} i\right)\ or\ \left(\frac{2}{3} -\frac{1}{3} \sqrt{5} i\right)[/tex]
