2022-01-11

Write a polynomial function of less the group for integral coefficients that has the given zero square root of two, 1+2 square root of two

nick1337

Remove parentheses.

$f\left(x\right)=\left(x-0\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x-\left(1+2\sqrt{2}\right)\right)\left(x+\left(1+2\sqrt{2}\right)\right)$

$f\left(x\right)=\left({x}^{2}+x\left(-\sqrt{2}\right)\right)\left(x+\sqrt{2}\right)\left(x-\left(1+2\sqrt{2}\right)\right)\left(x+1+2\sqrt{2}\right)$

Expand $\left({x}^{2}+2\left(-\sqrt{2}\right)\right)\left(x+\sqrt{2}\right)$ using the FOIL Method.

$f\left(x\right)=\left({x}^{2}x+{x}^{2}\sqrt{2}+x\left(-\sqrt{2}\right)x+x\left(-\sqrt{2}\right)\sqrt{2}\right)\left(x-\left(1+2\sqrt{2}\right)\right)\left(x+1+2\sqrt{2}\right)$

Simplify and combine like terms.

$f\left(x\right)=\left({x}^{3}-2x\right)\left(x-\left(1+2\sqrt{2}\right)\right)\left(x+1+2\sqrt{2}\right)$

Simplify each term.

$f\left(x\right)=\left({x}^{3}-2x\right)\left(x-1-2\sqrt{2}\right)\left(x+1+2\sqrt{2}\right)$

Expand $\left({x}^{3}-2x\right)\left(x-1-2\sqrt{2}\right)$ by multiplying each term in the first expression by each term in the second expression.

$f\left(x\right)=\left({x}^{3}x+{x}^{3}\cdot -1+{x}^{3}\left(-2\sqrt{2}\right)-2x\cdot x-2x\cdot -1-2x\left(-2\sqrt{2}\right)\right)\left(x+1+2\sqrt{2}\right)$

Simplify each term.

$f\left(x\right)=\left({x}^{4}-{x}^{3}+{x}^{3}\left(-2\sqrt{2}\right)-2{x}^{2}+2x+4x\sqrt{2}\right)\left(x+1+2\sqrt{2}\right)$

Expand $\left({x}^{4}-{x}^{3}+{x}^{3}\left(-2\sqrt{2}\right)-2{x}^{2}+2x+4x\sqrt{2}\right)\left(x+1+2\sqrt{2}\right)$ by multiplying each term in the first expression by each term in the second expression.

$f\left(x\right)={x}^{4}x+{x}^{4}\cdot 1+{x}^{4}\left(2\sqrt{2}\right)-{x}^{3}x-{x}^{3}\cdot 1-{x}^{3}\left(2\sqrt{2}\right)+{x}^{3}\left(-2\sqrt{2}\right)x+{x}^{3}\left(-2\sqrt{2}\right)\cdot 1+{x}^{3}\left(-2\sqrt{2}\right)\left(2\sqrt{2}\right)-2x2x$

$-2{x}^{2}\cdot 1-2{x}^{2}\left(2\sqrt{2}\right)+2x\cdot x+2x\cdot 1+2x\left(2\sqrt{2}\right)+4x\sqrt{2}x+4x\sqrt{2}\cdot 1+4x\sqrt{2}\left(2\sqrt{2}\right)$

Simplify terms.

$f\left(x\right)={x}^{5}-11{x}^{3}-2{x}^{3}\sqrt{2}+{x}^{3}\left(-2\sqrt{2}\right)+2x+4x\sqrt{2}+4x\sqrt{2}+16x$

Add $-2{x}^{3}\sqrt{2}$ and ${x}^{3}\left(-2\sqrt{2}\right)$

$f\left(x\right)={x}^{5}-11{x}^{3}-4{x}^{3}\sqrt{2}+2x+4x\sqrt{2}+4x\sqrt{2}+16x$

Add $2x$ and $16x$

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