Polynomials (1 Viewer)

[kaz1]

1.Express x⁴ + x² +1 as a difference of two squares. Hence factor x⁴ + x² +1 into two quadratic factors.

Figured out the first question.

2.Express x⁶ + 2x⁴ + 2x² +1 as a product of three quadratic factors.

I have no fucking clue.

 

[kaz1]

i

Thanks for bumping even though you were no help whatsoever.

 

[kaz1]

1. is an imaginary number.

2. can be done by dividing polynomials (x-1) as a factor

Since I am in year 11 we don't do complex numbers and x-1 is not a factor.

 

[hopethisworks]

lawl, my bad

 

[kaz1]

bump.

 

[12o9]

Since I am in year 11 we don't do complex numbers and x-1 is not a factor.

x²+1 is though

 

[ratcher0071]

1.Express x⁴ + x² +1 as a difference of two squares. Hence factor x⁴ + x² +1 into two quadratic factors.

Figured out the first question.

2.Express x⁶ + 2x⁴ + 2x² +1 as a product of three quadratic factors.

I have no fucking clue.

How can you express x^4 + x^2 +1 as difference of two squares?

 

[kaz1]

How can you express x^4 + x^2 +1 as difference of two squares?

x² + x⁴ +1
x²+ [(x²)² +2x² +1] -2x²
(x² + 1)² - x²

 

[lolokay]

1. x⁴ + 2x² + 1 - x²
= (x² + 1 - x)(x² + 1 + x)

2. x⁶ + 2x⁴ + x² + x² + 1
= (x² +1 )(x⁴ + x²) + (x² + 1)
= (x² +1 )(x⁴ + x² +1 )
= (x² + 1 - x)(x² + 1 + x)(x² +1 )

 

[fishy89sg]

oh man im bored..

2.Express x^6 + 2x^4 + 2x^2 +1 as a product of three quadratic factors.


let u = x^2

so your P(x) = u^3 + x^2 + u + 1
note that u = -1 is a root of P(x)
factorising that yields P(x) = (u+1)(u^2+u+1)
implying that P(x) = (x^2+1)(x^4+x^2+1)

 

[kaz1]

oh man im bored..

2.Express x^6 + 2x^4 + 2x^2 +1 as a product of three quadratic factors.


let u = x^2

so your P(x) = u^3 + x^2 + u + 1
note that u = -1 is a root of P(x)
factorising that yields P(x) = (u+1)(u^2+u+1)
implying that P(x) = (x^2+1)(x^4+x^2+1)

Thank you.