what are the real and imaginary zeros of the equation 2x^3 + 5x^2 - 3x + 7
I find no rational zeros, but the real zero is very near -1.3, so
f(x) = (x+1.3)(2x^2+6.4x-6.12)
I assume you have some graphical or numeric tools for solving general polynomials.
oops - the real root is at -3.282
To find the real and imaginary zeros of the equation 2x^3 + 5x^2 - 3x + 7, we will use a method called factoring by grouping.
Step 1: Factor out the greatest common factor (GCF) if it exists. In this case, there is no common factor other than 1.
Step 2: Group the terms in pairs and look for common factors among each pair. We have the equation:
(2x^3 + 5x^2) + (-3x + 7)
We can factor a common factor of x^2 from the first pair and a common factor of -1 from the second pair:
x^2(2x + 5) -1(3x - 7)
Step 3: Factor out the common factors from each group:
x^2(2x + 5) - 1(3x - 7)
Step 4: The resulting expression can be factored once more:
(x^2 - 1)(2x + 5) - (3x - 7)
Step 5: Notice that x^2 - 1 is a difference of squares, which can be further factored:
[(x - 1)(x + 1)](2x + 5) - (3x - 7)
Step 6: Distribute and simplify:
(2x - 2)(x + 1)(2x + 5) - (3x - 7)
Step 7: Simplify further:
(2x - 2)(x + 1)(2x + 5) - 3x + 7
Now we can solve for the zeros by setting the equation equal to zero:
(2x - 2)(x + 1)(2x + 5) - 3x + 7 = 0
The possible zeros are the values of x that make the equation equal to zero.
At this point, the equation becomes more difficult to solve using factoring alone. However, you can use numerical methods like the Newton-Raphson method or the synthetic division method to find approximate solutions.