To find all zeros of the polynomial [tex]\( 2x^2 - 3x^3 + 6x - 2 \)[/tex] given that two of its zeros are [tex]\(\sqrt{2}\)[/tex] and [tex]\(-\sqrt{2}\)[/tex], we can use the fact that polynomials can be factored into linear factors corresponding to each root.
Given the zeros [tex]\(\sqrt{2}\)[/tex] and [tex]\(-\sqrt{2}\)[/tex], the polynomial can be expressed as follows:
[tex]\[ f(x) = 2x^2 - 3x^3 + 6x - 2 \][/tex]
[tex]\[ f(x) = (x - \sqrt{2})(x + \sqrt{2})(Ax + B) = (x^2 - 2)(Ax + B) \][/tex]
Expanding the product [tex]\((x^2 - 2)(Ax + B)\)[/tex]:
[tex]\[ (x^2 - 2)(Ax + B) = A x^3 + B x^2 - 2 A x - 2 B \][/tex]
We need this expansion to match our original polynomial [tex]\( -3x^3 + 2x^2 + 6x - 2 \)[/tex]. Matching coefficients, we get:
[tex]\[-3x^3 + 2x^2 + 6x - 2 = A x^3 + B x^2 - 2 A x - 2 B \][/tex]
Now, let's equate the coefficients of the corresponding powers of [tex]\( x \)[/tex]:
For [tex]\( x^3 \)[/tex]:
[tex]\[ -3 = A \][/tex]
For [tex]\( x^2 \)[/tex]:
[tex]\[ 2 = B \][/tex]
For [tex]\( x^1 \)[/tex] (linear term):
[tex]\[ 6 = -2A \][/tex]
Substituting [tex]\( A = -3 \)[/tex] here, we can check if this holds true:
[tex]\[ 6 = -2(-3) \][/tex]
[tex]\[ 6 = 6 \, \text{(True)} \][/tex]
For the constant term:
[tex]\[ -2 = -2B \][/tex]
Substituting [tex]\( B = 2 \)[/tex] here, we can check if this holds true:
[tex]\[ -2 = -2(2) \][/tex]
[tex]\[ -2 = -4 \, \text{(True)} \][/tex]
From this, we find that the correct coefficients [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are:
[tex]\[ A = -3 \][/tex]
[tex]\[ B = 2 \][/tex]
So, the factorization of the polynomial is:
[tex]\[ f(x) = (x^2 - 2)(-3x + 2) \][/tex]
We can now solve for the remaining zeros by setting each factor to zero:
1. [tex]\( x^2 - 2 = 0 \)[/tex]
[tex]\[ x^2 = 2 \][/tex]
[tex]\[ x = \pm \sqrt{2} \][/tex]
These are the given zeros, [tex]\(\sqrt{2}\)[/tex] and [tex]\(-\sqrt{2}\)[/tex], which we already know.
2. [tex]\( -3x + 2 = 0 \)[/tex]
[tex]\[ -3x = -2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
Thus, the zeros of the polynomial [tex]\( 2x^2 - 3x^3 + 6x - 2 \)[/tex] are:
[tex]\[ \sqrt{2}, -\sqrt{2}, \text{ and } \frac{2}{3} \][/tex]
