Respuesta :
The first one factors easily. The first quadratic, I mean. The 2 numbers that add up to +2 and at the same time multiply to -15 and 5 and -3. So those are 2 of the 4 roots we have. The second quadratic does not factor so nicely. You need to put that into the quadratic formula to solve. [tex]x= \frac{-8+/- \sqrt{8^2-4(1)(17)} }{2} [/tex] which simplifies to [tex]x= \frac{-8+/- \sqrt{64-68} }{2} [/tex]. That gives us a negative radicand and that's a problem. [tex]x= \frac{-8+/- \sqrt{-4} }{2} [/tex]. Since -1 is equal to i^2, we can rewrite to begin dealing with the negative properly. [tex]x= \frac{-8+/- \sqrt{-1(4)} }{2} [/tex]. Replacing -1 with i^2 gives us [tex]x= \frac{-8+/- \sqrt{i^2(4)} }{2} [/tex]. i^2 has a perfect root of i in it, and 4 has a perfect square of 2 in it, so we simplify more to [tex]x= \frac{-8+/-2i}{2} [/tex]. The 2 in the denominator reduces with the numerator to give us a final 2 roots that are x = -4 + i, and x = -4 - i. Taking all those roots together, we find that the solution to our problem is choice B (although I believe you put some extra commas in there on accident).
Given polynomial function f(x) = (x^2 + 2x - 15)(x^2 + 8x + 17).
In order to find all roots of the given polynomial function, we need to set each of above factor equal to 0.
Therefore,
x^2 + 2x - 15 =0 and x^2 + 8x + 17 =0.
Let us solve first quadratic now.
x^2 + 2x - 15 =0
Factoring out quadratic
(x+5)(x-3) =0
x+5=0 or x-3=0
x=-5 or x=3.
Let us solve second quadratic equation now
x^2 + 8x + 17 =0.
Applying quadratic formula,
[tex]\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\cdot \:1\cdot \:17}}{2\cdot \:1}[/tex]
[tex]x=\frac{-8+\sqrt{8^2-4\cdot \:1\cdot \:17}}{2\cdot \:1}:\quad -4+i[/tex]
[tex]x=\frac{-8-\sqrt{8^2-4\cdot \:1\cdot \:17}}{2\cdot \:1}:\quad -4-i\[/tex]
[tex]x=-4+i,\:x=-4-i[/tex]
Therefore, the list of roots for the polynomial function is:
[tex]x=3,\:x=-5,\:x=-4+i,\:x=-4-i.[/tex]
