Answer to Question #285391 in Algebra for Alexander

Question

How would i solve| (x-2)(x²+8x-1).                    (X+4)(x²-5x+9), I have to find the answer to these, then add them together.

Answer 1

$(x-2)(x^2+8x-1)=0$

$x-2=0 \text{ or }x^2+8x-1=0$

$x-2=0=>x_1=2$

$x^2+8x-1=0$

$x^2+8x+16=17$

$(x+4)^2=17$

$x_2=-4-\sqrt{17}, x_3=-4+\sqrt{17}$

$\{-4-\sqrt{17},-4+\sqrt{17}, 2\}$

$(x+4)(x^2-5x+9)=0$

$x+4=0 \text{ or }x^2-5x+9=0$

$x+4=0=>x_4=-4$

$x^2-5x+9=0$

$D=(-5)^2-4(1)(9)=-11<0$

There are no real solutions.

$\{-4\}$

$-4-\sqrt{17}+(-4+\sqrt{17})+2+(-4)=-10$