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