sengihantq

2022-10-08

Find a polynomial of degree 3 having with the zeros $1+\sqrt{2},1-\sqrt{2}$, and 4.

selixaak

If a is a zero of polynomial, then (x-a) is its factor.
Given three zeros are $1+\sqrt{2},1-\sqrt{2},4$
The factors of polynomial are:
$\left(x-4\right),\left(x-\left(1+\sqrt{2}\right)\right),\left(x-\left(1-\sqrt{2}\right)\right)$
The polynomial of degree 3 will be of the form:
$f\left(x\right)=a\left(x-4\right)\left(x-\left(1+\sqrt{2}\right)\right)\left(x-\left(1-\sqrt{2}\right)\right)\phantom{\rule{0ex}{0ex}}=a\left(x-4\right)\left(x-1-\sqrt{2}\right)\right)\left(x-1+\sqrt{2}\right)\right)\phantom{\rule{0ex}{0ex}}=a\left(x-4\right)\left(\left(x-1\right)-\sqrt{2}\right)\right)\left(\left(x-1\right)+\sqrt{2}\right)\right)$
Then
$=a\left(x-4\right)\left(\left(x-1{\right)}^{2}-\left(\sqrt{2}{\right)}^{2}\right)\right)\phantom{\rule{0ex}{0ex}}=a\left(x-4\right)\left(\left(x-1{\right)}^{2}-2\right)\right)\phantom{\rule{0ex}{0ex}}=a\left(x-4\right)\left({x}^{2}-2x+1-2\right)\phantom{\rule{0ex}{0ex}}=a\left(x-4\right)\left({x}^{2}-2x-1\right)\phantom{\rule{0ex}{0ex}}=a\left(x\left({x}^{2}-2x-1\right)-4\left({x}^{2}-2x-1\right)\right)\phantom{\rule{0ex}{0ex}}=a\left(\left({x}^{3}-2{x}^{2}-x\right)-\left(4{x}^{2}-8x-4\right)\right)\phantom{\rule{0ex}{0ex}}=a\left({x}^{3}-2{x}^{2}-x-4{x}^{2}+8x+4\right)\phantom{\rule{0ex}{0ex}}=a\left({x}^{3}-6{x}^{2}+7x+4\right)$
Since the given polynomials in options have leading coefficient 1, we shall put a = 1.
$f\left(x\right)=a\left({x}^{3}-6{x}^{2}+7x+4\right)=1\left({x}^{3}-6{x}^{2}+7x+4\right)={x}^{3}-6{x}^{2}+7x+4\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{x}\mathbf{+}\mathbf{4}$

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