Question

$(x {}^{4} + 5x {}^{2} + 9)$

Answer 1

HI DUDE....GUD MRNG.......♥♥♥

given equation is x^4 + 5x^2 + 9

let x^2 =a

a^2 + 5a + 9=0

the equation comparing ax^2 + bx + c=0

a =1 , b = 5 ,c = 9

the formula is =( -b (+ or -) squ( b^2 - 4ac))/ 2a

= (-5 (+ or -) squ (5^2 -4*1*9))/2*1

= ( -5 ( +or -) squ(25 -36))/2

=( -5 (+ or -)squ ( -9))/2

(-5 + 3i) /2,( -5 -3i)/2 (since i^2= -1)

x^2= (-5 + 3i)/2 , x^2 = ( -5 -3i)/2

x = +or-squ ( -5 + 3i)/2 , x= + or -squ( -5 -3i)/2

the fractors are x = +or-squ ( -5 + 3i)/2 , x= + or -squ( -5 -3i)/2

HOPE THIS HELPS....♥♥♡♡

Answer 2

Heya! !
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$(x {}^{4} + 6x {}^{2} + 9) - x {}^{2} \\ = (x { }^{2} + 3) {}^{2} - x {}^{2} \\ = (x {}^{2} + 3 - x)(x {}^{2} + x + 3) \\ = (x {}^{2} + 3 - x)(x {}^{2} + x + 3)is \: the \: answer$