# Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.

Question
Polynomial arithmetic
Calculate: Polymomial equation with real coefficients taht has roots 3, 1 - i.

2021-03-10
Polynomial equation with real coefficients that has the roots 3, 1 -i
is $$x^{3} - 5x^{2} + 8x - 6 =0$$.
Given:
$$3, 1 - i$$
Formula Used: $$(a + b)(a - b) = a^{2} - b^{2}$$
Calculation:
If the polynomial has real coefficients, then it’s imaginary roots occur in conjugate pairs. So, a polynomial with the given root $$1 - i$$
must have another root as $$1 + i$$.
Since each root of the equation corresponds to a factor of the polynomial, also, the roots indicate zeros of that polynomial, thus, the polynomial equation is written as,
$$(x - 3) [x - (1 - i)] [x - (1 + i)] = 0$$
$$(x - 3) (x - 1 + i) (x - 1 - i) = 0$$
Further use arithmetic rule,
$$(a + b) (a - b) = a^{2} + b^{2}$$
$$\text{and} i^{2} = -1$$
Now, the polynomial equation is:
$$(x - 3)(x^{2} - 2x + 1 + 1) = 0$$
$$(x - 3)(x^{2} -2x +2) = 0$$
$$x^{3} - 3x^{2} - 2x^{2} + 6x + 2x - 6 = 0$$
$$x^{3} - 5x^{2} = 8x - 6 = 0$$
Hence, the polynomial equation of given roots $$3, 1 - i \text{is} x^{3} - 5x^{2} + 8x - 6 = 0$$

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Given the following function: $$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}-{1.99}$$ a)Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$ b)Redo the same calculation by first rewriting the equation using the polynomial factoring technique c)Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
Given the following function:
$$\displaystyle{f{{\left({x}\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}$$
a) Use three-digit rounding frithmetic, the assumption that $$\displaystyle{e}^{{{1.53}}}={4.62}$$, and the fact that $$\displaystyle{e}^{{{n}{x}}}={\left({e}^{{{x}}}\right)}^{{{n}}}$$ to evaluate $$\displaystyle{f{{\left({1.53}\right)}}}$$
b) Redo the same calculation by first rewriting the equation using the polynomial factoring technique
c) Calculate the percentage relative errors in both part a) and b) to the true result $$\displaystyle{f{{\left({1.53}\right)}}}=-{7.60787}$$
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