f(x) = x^3+6x^2 +5x−12
find all possible zeros. find all real zeros using synthetic division, factoring, and the quadratic formula.
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POLYNOMIALS QUADRATIC FORMULA FACTORING POLYNOMIALS REAL ROOTS
K w. asked • 10/11/17
Find all real roots of the given polynomials, using the quadratic formula if necessary. Then write the polynomial in factored form.
a) f(x) = x3 − 5x2 + 2x + 12
b) g(x) = 4x3 − 6x2 + 1
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Andrew M. answered • 10/11/17
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Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
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a) f(x) = x3 − 5x2 + 2x + 12
Since this is a 3rd degree polynomial, we expect 3 roots.
The possible rational roots are ±(factors of the constant... 12)/(factors of leading coefficient... 1)
Possible rational roots: ±(factors of 12)/(factors of 1)
±{1, 2, 3, 4, 6, 12}/1
Using synthetic division find look for a root:
If x = 1 is a zero then (x-1) is a root
1 | 1 -5 2 12
| 1 -4 -2
-------------------------
1 -4 -2 | 10 Since remainder isn't zero, not a root
------
If x=3 is a zero, then (x-3) is a root
3 | 1 -5 2 12
| 3 -6 -12
-------------------------
1 -2 -4 | 0 Since remainder is zero, this is a root
----
x3 − 5x2 + 2x + 12 = (x-3)(x2 - 2x - 4)
Now we need to solve or factor x2-2x -4
This isn't factorable. We can continue synthetic division
or use quadratic equation.
x = [2 ±√((-2)2-4(1)(-4))]/2 =
= (2 ±√20)/2
= (2 ±2√5)/2
= 1±√5
We now have 3 roots:
x = {3, 1+√5, 1 - √5}
f(x) = (x-3)(x - (1+√5))(x-(1-√5))
= (x-3)(x - 1 - √5)(x - 1 + √5)
b) g(x) = 4x3 − 6x2 + 1
Again we would expect 3 roots since highest exponent is 3.
Possible rational roots: ±1/{1, 2, 4} = ±(1, 1/2, 1/4)
Again.. Look for a root using synthetic division.
After finding the first root you can use quadratic on the rest or,
if possible, factor it.
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