Question

Find 'a' if (x+a) is a factor of a x^4- a^2x² + 3x -a.​

Answer 1

Answer:

$\mathrm{solve\:for\:a,\:x^4-a^2x^2+3x-a=x+a\quad :\quad a=-\frac{1+\sqrt{x^3\left(x^3+2\right)+1}}{x^2},\:a=-\frac{1-\sqrt{x^3\left(x^3+2\right)+1}}{x^2};\neq 0}$

Step-by-step explanation:

$x^4-a^2x^2+3x-a=x+a$

$\mathrm{Subtract\:}a\mathrm{\:from\:both\:sides}$

$x^4-a^2x^2+3x-a-a=x+a-a$

$x^4-a^2x^2+3x-2a=x$

$\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}$

$x^4-a^2x^2+3x-2a-x=x-x$

$-x^2a^2-2a+x^4+2x=0$

$a_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\left(-x^2\right)\left(x^4+2x\right)}}{2\left(-x^2\right)}$

$a_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{1+x^3\left(x^3+2\right)}}{2\left(-x^2\right)};\quad \:x\ne \:0$

$\mathrm{Separate\:the\:solutions}$

$a_1=\frac{-\left(-2\right)+2\sqrt{1+x^3\left(x^3+2\right)}}{2\left(-x^2\right)},\:a_2=\frac{-\left(-2\right)-2\sqrt{1+x^3\left(x^3+2\right)}}{2\left(-x^2\right)}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$a=-\frac{1+\sqrt{x^3\left(x^3+2\right)+1}}{x^2},\:a=-\frac{1-\sqrt{x^3\left(x^3+2\right)+1}}{x^2};\quad \:x\ne \:0$