Given a zero: x=2, of the polynomial: x3-4x2+21x-34=0, find the others.?

Answer 1

If you're given a zero at x=2, that means that one factor of the polynomial is (x-2). Just divide the polynomial by (x-2), and in this case the result will be a quadratic that you can solve for the other roots.

(x^3 - 4x^2 + 21x - 34) / (x - 2)

It's done like long division. Take the term with highest degree of x in the numerator (x^3) and divide by the the term with highest degree of x in the denominator (x).

x^3 / x = x^2

That's the first part of your solution. Now multiply that (x^2) by the denominator (x - 2)

x^2 (x - 2) =
x^3 - 2x^2

And subtract that from the numerator:

x^3 - 4x^2 + 21x - 34 - (x^3 - 2x^2) =
x^3 - x^3 - 4x^2 + 2x^2 + 21x - 34 =
-2x^2 + 21x - 34

Take the term with highest degree of x in what is left of the numerator (-2x^2) and divide by the the term with highest degree of x in the denominator (x).

-2x^2 / x = -2x

That's the next part of your solution. Now multiply that (-2x) by the denominator (x - 2)

-2x (x - 2) =
-2x^2 + 4x

And subtract that from what is left of the numerator:

-2x^2 + 21x - 34 - (-2x^2 + 4x) =
-2x^2 + 2x^2 + 21x - 4x - 34 =
17x - 34

And that is 17 times the denominator, so the final part of the solution is 17.

So, the equation factors as:

(x - 2)(x^2 -2x + 17)

The discriminant of that quadratic (2^2 - 4*17 = -64) is negative, so the other two roots are not real. In case you want them in imaginary numbers, they are:

x = (2 +/- sqrt((-2)^2 - 4*1*17))/(2*1)
x = (2 +/- sqrt(-64)) / 2
x = 1 +/- 4i

The roots are 1 + 4i and 1 - 4i, so the equation has no (real) zeroes aside from x=2.

Answer 2

34 = 2*17 (x^3 - 4*x^2 + 21*x - 34) = 0 (x^3 - 4*x^2 + 21*x - 34)/(x - 2) = x^2 - 2*x + 17 x^2 - 2*x + 17 is prime. solve(x^2 - 2*x + 17) x = 1 + 4*i, x = 1 - 4*i Answer: x1 = 2, x2 = 1 + 4*i, x3 = 1 - 4*i

Answer 3

You don't even need x=2, just plug the whole thing into your graphing calculator and find where the graph intercepts the x axis. And for future reference, use x^3 instead of x3, otherwise it just looks like multiplication.

Answer 4

By solvin we get 1+ i4 and 1-i4 (complex roots)

solution :
as one root is x=2 ,
(x3-4x2+21x-34) / (x-2) gives the other two roots.

on solving that we get x2-2x+17 as the eqn (i.e., (x-2)(x2-2x+17)= (x3-4x2+21x-34) )

and to get the two roots we need to solve the quadratic eqn(x2-2x+17) as it has the complex roots ,by solving we get
1+i4 and 1-i4.

for ur reference :
i) The quadratic eqn is obtained by synthetic division method.
i.e)
x2 - 2x + 17 (quotient)
**** ________________
x-2 | x3 - 4x2 + 21x - 34
****** x3 - 2x2 (change sign and cancel first term)
*******-------------
********* - 2x2 + 21x
********* - 2x2 + 4x (change sign and cancel 1st term)
**********---------------
******************17x - 34
******************17x - 34 (change sign and cancel 1st term)
****************** -----------
********************* 0

[jz to make it look like division i hv used '*' ( as i was not able to leave blank spaces) , dont get confused :-) ]

ii) quadratic eqn(of the model ax2+bx+c=0) is solved using the formula
[-b (+or-) sqrt(b2 - 4 a.c)] / 2a
(a,b and c are the co-eff of x2,x and constant term resp.)