How do I find Real or Complex Zeros?

Question

Find all real or complex zeros:

f(x) = 3x^4 + 11x^3 + 17x^2 + 29x - 12

How? Thank you.

Answer 1

If you can find one root, then you can factor. With a little trial and error, you should find x = -3 is a real root.

f(-3) = 3(-3)^4 + 11(-3)^3 + 17(-3)² + 29(-3) - 12
f(-3) = 3(81) + 11(-27) + 17(9) + 29(-3) - 12
f(-3) = 243 - 297 + 153 - 87 - 12
f(-3) = 0

Okay, so factor out (x + 3) using synthetic division.

f(x) = (x+3)(3x^3 + 2x² + 11x - 4)

What will work in this new function?
g(x) = 3x^3 + 2x² + 11x - 4

More trial and error to find x = 1/3
g(1/3) = 3(1/3)^3 + 2(1/3)² + 11(1/3) - 4
g(1/3) = 1/9 + 2/9 + 11/3 - 4
g(1/3) = 1/3 + 11/3 - 4
g(1/3) = 4 - 4 = 0

x - 1/3 would be a factor, but I'm going to multiply it by 3 to get rid of the fraction. So we have 3x - 1. Again do synthetic division:

f(x) = (x - 3)(3x - 1)(x² + x + 4)

Now you can use the quadratic formula on this last part.

h(x) = x² + x + 4
a = 1, b = 1, c = 4

...... -b ± sqrt( b² - 4ac )
x = ---------------------------
................... 2a

...... -1 ± sqrt( 1² - 4(1)(4) )
x = ---------------------------
................... 2(1)

...... -1 ± sqrt( -15 )
x = ---------------------
................... 2

...... -1 ± sqrt(5)i
x = -----------------
................... 2

So the roots are:
Real roots:
-3
1/3
Complex roots:
-1/2 + (sqrt(5)/2) i
-1/2 - (sqrt(5)/2) i


Edit:
Kudos to the other answerer for mentioning how to figure out possible real roots. I completely left that out.

Indeed rather than just guessing, you can restrict yourself to factors of the coefficient on the last term divided by factors of the coefficient on the first term.

Factors of -12 are ±1, ±2, ±3, ±4, ±6 and ±12
Factors of 3 are ±1 and ±3.
So the complete set to try is:
{±1/3, ±2/3, ±4/3, ±1, ±2, ±3, ±4, ±6, ±12}

At the second step you reduce this set further to:
{±1/3, ±2/3, ±4/3, ±1, ±2, ±4}

Answer 2

Finding the zeroes for polynomials requires you to factor the equation so all of the factors have degree 1. (x without an exponent) The first step is to find all the factors of the constant at the end -12 and the coefficient of the term with the largest power. Then all of the zeroes are each factor of -12 over each factor of 3. so possibilities are 4, -4, 12, -12, 2, -2, 6, -6, 1, -1, 2/3, -2/3, 1/3, and -1/3. Then you have to try all of these and see which ones work. Other ways include factoring the polynomial on your calculator (only TI-89 and above), or graphing the functions and going to calculate tab and select zero. To find the imaginary zeroes you must first factor the polynomial into 2nd degree and first degree polynomials. Then use the quadratic formula. factorization is (x+3)(3x-1)(x^2+x+4) or (x+3)(3x-1)(x+.5-(15^(1/2)*i)/2)(x+.5+(15^(1/2)*i)/2)

Answer 3

f(x) = 7x² + 3x - a million discriminant: D = 3² - 4*7*(-a million) = 35 > 0 f(x) has 2 genuine zeros and no complicated 0 made from the zeros: x1*x2 = c/a = -a million/7 < 0 The zeros of f(x) are of opposite signs and indications. One is valuable, the different is damaging

Answer 4

One way to do it is to graph the function and find all the places it crosses the x axis.
Another way to do it is to factor your function into as many separate factors as you can and set each equal to zero. Then solve each for x.