Question

factorize the expression x⁴+324​

Answer 1

Answer:

$( x + 3\sqrt{2} ) ( x - 3\sqrt{2}  )  (x^2 + 18)$

Step-by-step explanation:

Given :

To factorize the expression $x^4 + 324$

Solution :

We know that,

( a + b )² = a² + 2ab + b²  ..(i)

a² - b² = ( a + b ) ( a - b )  ..(ii)

The expression,

⇒ $x^4 + 324$

⇒ $x^4 + 36 x^2 - 36x^2 + 18^2$

⇒ $x^4 + 36 x^2 + 18^2 - 36x^2$

⇒ $[x^4 + 2(18)(x^2) + 18^2] - 36x^2$

By using identity (i),

where a = x² , b = 18,

We get,

⇒ $(x^2 + 18)^2 - 36x^2$

⇒ $(x^2 + 18)^2 - (6x)^2$

By using identity (ii),

We get,

⇒ ( x² - 6x + 18 ) ( x² + 6x + 18 )

Answer 2

Concept:

Algebraic Identities

$(x^2-y^2 )=(x-y)^* (x+y)\\\\(x+y)^2=x^2+2xy+y^2$

Given:

x⁴+324​

Find:
Factors

Solution:

We need to manipulate the given polynomial such that it can be expressed in the form of the product of two or more polynomials.

$(x^2 )^2+(18)^2$

Both of the terms are perfect squares so if we add and subtract 2*18*x^2 to it then we can use the identity $(x+y)^2=x^2+2xy+y^2$ on it.

$x^4+36x^2-36x^2+18^2\\\\(x^2 )^2+2(18)x+(18)^2-36x^2\\\\(x^2+18)^2-(6x)^2$

now using the identity $(x^2-y^2 )=(x-y)^* (x+y)$ gives

$(x^2+18-6x)(x^2+18+6x)$

Now both of these are standard quadratic equations and can be factored

using quadratic formula into $((-3 - 3 i) + x) ((-3 + 3 i) + x) ((3 - 3 i) + x) ((3 + 3 i) + x)$

but these factors are complex and involve imaginary terms with iota.

Hence the factors of $x^4+324$ are $(x^2+18-6x)(x^2+18+6x)$

#SPJ2