Question 824327
A polynomial of degree 3 should have 3 zeros. But only two have been given. The third zero must be found in order to find the polynomial.<br>
The key to finding the third zero is: If a polynomial with real coefficients has complex zeros, then they will always come in conjugate pairs. Since we were given one complex zero, 2 - i, then the missing zero must be its conjugate: 2 + i.<br>
So the three zeros are -2, 2 - i and 2 + i. And when a number, let's call it "z", is a zero of a polynomial, then (x - z) is a factor of the polynomial. So, in factored form,
{{{f(x) = (x - (-2))(x - (2-i))(x-(2+i))}}}
Simplifying each factor we get:
{{{f(x) = (x + 2)(x - 2+i)(x-2-i)}}}<br>
All that is left is to multiply this out. (Hint for future problems: Multiply the factors with complex conjugate zeros together first.) Multiplying the last two factors can be done with a clever use of the {{{(a+b)(a-b) = a^2-b^2}}}. To show you how I will do some grouping within those factors:
{{{f(x) = (x + 2)((x - 2)+i)((x-2)-i)}}}
Treating the "(x-2)" as the "a" of the pattern and the "i" as the "b", this pattern tells us that multiplying the last two factors will result in {{{a^2-b^2}}}:
{{{f(x) = (x + 2)((x - 2)^2-i^2)}}}
We can use another pattern, {{{(a-b)^2 = a^2-2ab+b^2}}} to square the (x-2). And {{{i^2 = -1}}}. So we get:
{{{f(x) = (x + 2)((x)^2 - 2(x)(2)+(2^2)-(-1))}}}
which simplifies as follows:
{{{f(x) = (x + 2)(x^2 - 4x+4+1)}}}
{{{f(x) = (x + 2)(x^2 - 4x+5)}}}<br>
Last we multiply the remaining factors (which I will leave to you). Just multiply each term of (x+2) times each term of {{{(x^2 - 4x+5)}}}. (That's 6 multiplications!) Then add any like terms. Then put the terms in standard (highest exponent to lowest exponent) order.