"Find all real solutions of" basically means solve for x. You're looking for the real numbers x such that the given equation is satisfied. "Zeros" is synonymous with "solutions".
It looks like the first equation is
3x³ + x² - 38x + 24 = 0
It's not immediately obvious whether this can be factorized, so we first carry out the rational zero test to see if there are any rational solutions. *If* this equation has any rational solutions, then they must take the form of (divisor of 24)/(divisor of 3). That is, take any divisor of the constant coefficient and divide it by any divisor of the leading term's coefficient.
We have
• divisors of 24 : ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24
• divisors of 3 : ±1, ±3
Sadly, there are 64 possible roots to test, but we only need one to make the rest of the problem easier.
• Let x = +1 / +1 = 1. Then plugging x = 1 into the equation gives
3•1³ + 1² - 38•1 + 24 = -10 ≠ 0
so x = 1 is *not* a solution
• Let x = +2 / +1 = 2. If x = 2, then
3•2³ + 2² - 38•2 + 24 = -24 ≠ 0
so x = 2 is *not* a solution.
• Let x = +3 / +1 = 3. If x = 3, then
3•3³ + 3² - 38•3 + 24 = 0
so x = 3 is a solution.
By the remainder theorem, this means that x - 3 divides 3x³ + x² - 38x + 24. Using long or synthetic division,
(3x³ + x² - 38x + 24) / (x - 3) = 3x² + 10x - 8
so that
3x³ + x² - 38x + 24 = 0
is the same as
(x - 3) (3x² + 10x - 8) = 0
We factorize the remaining quadratic to get
3x² + 10x - 8 = (3x - 2) (x + 4)
and so the original equation is equivalent to
(x - 3) (3x - 2) (x + 4) = 0
Solve for x :
x - 3 = 0 or 3x - 2 = 0 or x + 4 = 0
x = 3 or x = 2/3 or x = -4
The other equations can be solved similarly, though some of the others you included are much simpler.
For the second problem (note that you left out an exponent), we solve the equation
x³ - 2x² - 23x + 60 = 0
Again, not immediately obvious how to factorize this. But using the rational root test, we have
• divisors of 60 : ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60
• divisors of 1 : ±1
If you follow the same steps as before, you'll find the first rational solution with x = 3. So x - 3 divides x³ - 2x² - 23x + 60, and long division gives
(x³ - 2x² - 23x + 60) / (x - 3) = x² + x - 20
and the resulting quadratic is easily factored,
x² + x - 20 = (x - 4) (x + 5)
So, we have
x³ - 2x² - 23x + 60 = (x - 3) (x - 4) (x + 5) = 0
which means
x - 3 = 0 or x - 4 = 0 or x + 5 = 0
x = 3 or x = 4 or x = -5
For the third problem, we can factor by grouping.
x³ - 4x² - x + 4 = 0
x² (x - 4) - (x - 4) = 0
(x² - 1) (x - 4) = 0
(x - 1) (x + 1) (x - 4) = 0
Then
x - 1 = 0 or x + 1 = 0 or x - 4 = 0
x = 1 or x = -1 or x = 4
For the fourth problem, we use the rational root test again, but twice this time.
x⁴ - 6x³ + 7x² + 6x - 8 = 0
• divisors of -8 : ±1, ±2, ±4, ±8
• divisors of 1 : ±1
We find that x = 1 and x = 2 are both rational roots, so both x - 1 and x - 2 divide x⁴ - 6x³ + 7x² + 6x - 8. By long division,
(x⁴ - 6x³ + 7x² + 6x - 8) / ((x - 1) (x - 2)) = x² - 3x - 4
and factoring this gives
x² - 3x - 4 = (x + 1) (x - 4)
So, we have
x⁴ - 6x³ + 7x² + 6x - 8 = (x - 1) (x - 2) (x + 1) (x - 4) = 0
which means
x - 1 = 0 or x - 2 = 0 or x + 1 = 0 or x - 4 = 0
x = 1 or x = 2 or x = -1 or x = 4
For the last problem,
x⁴ + 4x³ + 7x² + 16x + 12 = 0
we use the rational root test again.
• divisors of 12 : ±1, ±2, ±3, ±4, ±6, ±12
• divisors of 1 : ±1
We find that both x = -1 and x = -3 are rational solutions, and dividing gives
(x⁴ + 4x³ + 7x² + 16x + 12) / ((x + 1) (x + 3)) = x² + 4
which we cannot factorize further over the real numbers. So we end up with
x⁴ + 4x³ + 7x² + 16x + 12 = (x + 1) (x + 3) (x² + 4) = 0
so that
x + 1 = 0 or x + 3 = 0 or x² + 4 = 0
x = -1 or x = -3
(We omit the third equation here because it has no real solution since x² + 4 ≥ 4 for all real x.)
