Question

Find the real or imaginary solutions of each polynomial equation by factoring.

Answer 1

I want to know If I am correct.
1) 2x+4x^2-30x=0 x=0 x=7
2) x^3+27=0 x=-3
3) x^4-13x^2+36=0 x=3, -3 , 2 , -2
4) 8x^3-1=0 x=1/2
5) x^4+21x^2-100=0 x=2, -2 , 5i, -5i
6) 2x^3+3x^2-8X-12=0 x=2, -2 , -3/2
Are these correct?

Answer 2

#1 is correct

#2 is partially correct...you have the real solution,but you are missing the two other imaginary solutions

Answer 3

three is right i believe (:

Answer 4

#3 is correct

#4, again, you only found the real solution and you're missing the 2 imaginary solutions

Answer 5

#5 is correct

Answer 6

#6 is correct

Answer 7

Could you tell me what I did wrong with the ones I am wrong?

Answer 8

with the ones you got wrong, you isolated x like normal, but you were supposed to use the sum of cubes factoring rule for #2 and the difference of cubes factoring rule for #4

Answer 9

sum of cubes factoring rule

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

Answer 10

difference of cubes factoring rule

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Answer 11

Could you help me out with #2? @jim_thompson5910
That way I could do the other one by myself.

Answer 12

27 = 3^3, so that means we have a sum of cubes


x^3+27=0

x^3+3^3=0

(x+3)(x^2 - 3*x + 3^2) = 0

(x+3)(x^2 - 3x + 9) = 0

x+3 = 0 or x^2 - 3x + 9 = 0 ... use zero product property here

x = -3 or x^2 - 3x + 9 = 0

So you can see that x = -3 is one of the solutions, but there are 2 more and they are found by solving x^2 - 3x + 9 = 0

Answer 13

To solve \(\large x^2 - 3x + 9 = 0\), we use the quadratic formula

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

\[\Large x = \frac{-(-3)\pm\sqrt{(-3)^2-4(1)(9)}}{2(1)}\]

\[\Large x = \frac{3\pm\sqrt{9-(36)}}{2}\]

\[\Large x = \frac{3\pm\sqrt{-27}}{2}\]

\[\Large x = \frac{3+\sqrt{-27}}{2} \ \text{or} \ x = \frac{3-\sqrt{-27}}{2}\]

\[\Large x = \frac{3+\sqrt{-1*9*3}}{2} \ \text{or} \ x = \frac{3-\sqrt{-1*9*3}}{2}\]

\[\Large x = \frac{3+\sqrt{-1}*\sqrt{9}*\sqrt{3}}{2} \ \text{or} \ x = \frac{3-\sqrt{-1}*\sqrt{9}*\sqrt{3}}{2}\]

\[\Large x = \frac{3+i*3*\sqrt{3}}{2} \ \text{or} \ x = \frac{3-i*3*\sqrt{3}}{2}\]

\[\Large x = \frac{3+3i\sqrt{3}}{2} \ \text{or} \ x = \frac{3-3i\sqrt{3}}{2}\]