sengihantq

2022-10-08

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

selixaak

Beginner2022-10-09Added 6 answers

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:

$$(x-4),(x-(1+\sqrt{2})),(x-(1-\sqrt{2}))$$

The polynomial of degree 3 will be of the form:

$$f(x)=a(x-4)(x-(1+\sqrt{2}))(x-(1-\sqrt{2}))\phantom{\rule{0ex}{0ex}}=a(x-4)(x-1-\sqrt{2}))(x-1+\sqrt{2}))\phantom{\rule{0ex}{0ex}}=a(x-4)((x-1)-\sqrt{2}))((x-1)+\sqrt{2}))$$

Then

$$=a(x-4)((x-1{)}^{2}-(\sqrt{2}{)}^{2}))\phantom{\rule{0ex}{0ex}}=a(x-4)((x-1{)}^{2}-2))\phantom{\rule{0ex}{0ex}}=a(x-4)({x}^{2}-2x+1-2)\phantom{\rule{0ex}{0ex}}=a(x-4)({x}^{2}-2x-1)\phantom{\rule{0ex}{0ex}}=a(x({x}^{2}-2x-1)-4({x}^{2}-2x-1))\phantom{\rule{0ex}{0ex}}=a(({x}^{3}-2{x}^{2}-x)-(4{x}^{2}-8x-4))\phantom{\rule{0ex}{0ex}}=a({x}^{3}-2{x}^{2}-x-4{x}^{2}+8x+4)\phantom{\rule{0ex}{0ex}}=a({x}^{3}-6{x}^{2}+7x+4)$$

Since the given polynomials in options have leading coefficient 1, we shall put a = 1.

$$f(x)=a({x}^{3}-6{x}^{2}+7x+4)=1({x}^{3}-6{x}^{2}+7x+4)={x}^{3}-6{x}^{2}+7x+4\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{)}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{x}\mathbf{+}\mathbf{4}$$

Given three zeros are $$1+\sqrt{2},1-\sqrt{2},4$$

The factors of polynomial are:

$$(x-4),(x-(1+\sqrt{2})),(x-(1-\sqrt{2}))$$

The polynomial of degree 3 will be of the form:

$$f(x)=a(x-4)(x-(1+\sqrt{2}))(x-(1-\sqrt{2}))\phantom{\rule{0ex}{0ex}}=a(x-4)(x-1-\sqrt{2}))(x-1+\sqrt{2}))\phantom{\rule{0ex}{0ex}}=a(x-4)((x-1)-\sqrt{2}))((x-1)+\sqrt{2}))$$

Then

$$=a(x-4)((x-1{)}^{2}-(\sqrt{2}{)}^{2}))\phantom{\rule{0ex}{0ex}}=a(x-4)((x-1{)}^{2}-2))\phantom{\rule{0ex}{0ex}}=a(x-4)({x}^{2}-2x+1-2)\phantom{\rule{0ex}{0ex}}=a(x-4)({x}^{2}-2x-1)\phantom{\rule{0ex}{0ex}}=a(x({x}^{2}-2x-1)-4({x}^{2}-2x-1))\phantom{\rule{0ex}{0ex}}=a(({x}^{3}-2{x}^{2}-x)-(4{x}^{2}-8x-4))\phantom{\rule{0ex}{0ex}}=a({x}^{3}-2{x}^{2}-x-4{x}^{2}+8x+4)\phantom{\rule{0ex}{0ex}}=a({x}^{3}-6{x}^{2}+7x+4)$$

Since the given polynomials in options have leading coefficient 1, we shall put a = 1.

$$f(x)=a({x}^{3}-6{x}^{2}+7x+4)=1({x}^{3}-6{x}^{2}+7x+4)={x}^{3}-6{x}^{2}+7x+4\phantom{\rule{0ex}{0ex}}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{)}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{x}\mathbf{+}\mathbf{4}$$

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