Solve x^3 + 2x^2 -13x +10 = 0 without squaring it.?

Question

Solve x^3 + 2x^2 -13x +10 = 0

I know that there needs to be 3 values for x however, 2 are not real (x crosses at only 1 point around -1 and -.8)

Should I use the quadratic formula, sythetic division, or the rational zero test.

Please help and solve

Answer 1

The sum of all of the coefficients gives you the value at x=1, and in this case, that sum is 0, so x=1 is a root.

Divide by x-1: First, divide the leading terms, x^3/x=x^2. So x^2 is the first term in the coefficient.

Get the remainder after dividing by x^2:
x^3+2x^2-13x+10 - x^2(x-1) = 3x^2-13x+10.

Divide leading terms again: 3x^2/x=3x, so 3x is the second term in the quotient.

Get the remainder:
3x^2-13x+10 - 3x(x-1) = -10x+10.

Divide leading terms again: -10x/x=-10, so -10 is the third term in the quotient.

Get the remainder:
-10x+10-(-10)(x-1) = 0, as expected, since we know (x-1) is a factor of x^3 + 2x^2 - 13x + 10.

Therefore, x^3 + 2x^2 - 13x + 10 = (x-1)(x^2+3x-10).

Now, factor x^2+3x-10 to get the other two roots. It factors nicely. You can always fall back on the quadratic formula.

Answer 2

Whato do you mean "2 are not real"? This function not only has three real solutions, but in fact has three _integer_ solutions. Anyway, I'd use synthetic division to test the possible roots given by the rational roots theorem, then once I found one, I'd complete the square on the remaining quadratic to find the other two roots.

Answer 3

Find the real root, factorize by synthetic division as x minus the real root and use quadratic formula with the other quadratic equation, to end up with 3 roots

Answer 4

firstly, you could try graphing the equation on your calculator and finding the intercepts.
all three solutions are real -- in fact they are all integers!
it is easy to see that x=1 is a solution to the equation
you could try synthetic division to obtain the quadratic and then solve the quadratic by factoring or using the quadratic formula
hope this helps!

Answer 5

we can see that the sum the coeffecients is eqal to zero. therefore x= +1 is one of the roots . So the equation can be written as (x-1)(ax^2+bx+c). Use synthetic division to find the coeffecients a,b,c and solve the quadratic equation for x .

Answer 6

let x=1 then poly. = 0 so x-1 is a factor thus divide poly. by x-1 using long division to reduce to x-1 times a quadratic factor

(x-1)(x^2+3x-10)

factor to (x-1)(x+5)(x-3)

set each factor = 0 solve for x to get x={1,-5,3}

Answer 7

Try googling "cubic equation solve analytically". Gives you a simple walkthrough - link below.

On the other hand, you could use your brain. Just looking at it I can see a real root at x=+1.

Answer 8

I'm not sure how to do this problem due to the 4 numbers instead of the normal 3. So here are some sites that you may use as helpful.

http://www.1728.com/quadratc.htm
http://www.mste.uiuc.edu/exner/ncsa/quad/
http://www.purplemath.com/modules/quadform.htm

Answer 9

OR, you could always plug it in a graphing calculator and find the x-int

Answer 10

Be happy to,except that I/we solving it for you,won't help you in the long run