Solve X3-2x2-14x+40 given that 3+I is a root?

Answer 1

Lots of great answers which deserve the 10 points, but a short-cut would be to use
product of roots = -40

The product of (3 + i) and (3 - i) is 10, so the other root must be -4.

Of course, the advantage of the other methods is that they check up that the given information is correct, whereas this method accepts the information that 3 + i is a root (so that 3 - i must be, too).

Answer 2

If 3+i is a root then 3-i must also be a root:(x-3-i)*(x-3+i)=x^2-6x+10 must be a factor of x^3-2x^2-14x+40. By dividing I get x+4 as the quotient. Therefore, x=-4.

Answer 3

since 3+i is a root of the equation, 3-i must also be a root of the equation (according to a theorem from algebra). two factors must therefore be (x-(3+i)) and (x-(3-i)). if you multiply these two together, you get (x-(3+i)) * (x-(3-i)) = x^2 -x(3+i) -x(3-i) +(3+i)* (3-i) =x^2 -3x -ix -3x +ix +9 -i^2 =x^2 -12x +9 +1 =x^2 -12x +10
you know that x^2 -6x +10 must be a factor of the equation, so you can use long division to find the remaining factor:
X3- 2x2 -14x +40 /(x^2 -6x +10) =x +4
the roots are therefor -4, 3+i, and 3-i

Answer 4

If 3+i is a root, then 3-i is a root too.
Then you write the two roots as factors of the polynom:
(x - (3+i))(x - (3-i)) = x^2 - x(3-i) - x(3+i) + (3+i)(3-i)
= x^2 + x(-3+i-3-i) + 9 - i^2 = x^2 - 6x + 10

Now you divide your polynom by x^2 - 6x + 10 and you will be left with a first order equation in x.

Answer 5

If 3 + i is a root, then 3 - i will be a root.

(x - (3+i))(x - (3-i))
= x² - (3 + i)x - (3 - i)x + (3+i)(3-i)
= x² -6x -ix + ix + 3² -3i + 3i -i²
= x² - 6x + 9 -(-1)
= x² - 6x + 10

So divide x^3 - 2x² - 14x + 40 by x² - 6x + 10, to get the 3rd root.

You can use synthetic division:
........... .............1 . 4
1 -6 10 ) 1 -2 -14 40
............. 1 -6 . 10
............. --------------
................. 4 -24 40
................. 4 -24 40
................ ------------
................ ............ 0

So you get (x + 4)

Your final factored result is:
(x + 4) (x - (3+i)) (x - (3-i)) = 0

And your roots are:
x = -4
x = 3 + i
x = 3 - i

Answer 6

if 3+i is a root then so is 3-i

(x-(3+i))(x-(3-i))
x^2-(3-i)x-(3+i)x+(9-i^2)
x^2-3x+ix-3x-ix+10
x^2-6x+10

divide that into the whole to get your answer the third factor
from that you can get your root
this is hard to represent by typing so I'll just tell you that I got
x+4 as the factor
so x=-4 is the final root

Answer 7

answer x= a million, 5, 8 Given x= 5 == > x-5 is a component component out (x-5) x^3 -14x^2 + 53x -40 x^2 <=== x^3-5x^2 { x^2(x-5)} -9x^2 + 53x -40 -9x <=== -9x^2+45x { 9x(x-5)} 8x -40 8 <=== 8x-40 { 8(x-5)} 0___ x^2 - 9x + 8 x^3 -14x^2 + 53x -40 =(x-5)(x^2-9x+8) =(x-5)(x-8)(x-a million)

Answer 8

yes

Answer 9

do your own homework