Question

Please help me !!

HOW TO FURTHER SOLVE IT.

Answer 1

How can we write this :-

x^4 + p x^2 + q = (x^2 + 2 x + 5) (x^2 + a x + b)

Answer 2

@hartnn @jim_thompson5910

Answer 3

you want to get a,b,p,q values ?

Answer 4

foil on right side,
(x^2 + 2 x + 5) (x^2 + a x + b) = x^4+.... ?

Answer 5

Assuming x^2 + 2x + 5 is a factor is x^4 + px^2 + q, then it follows that

x^4 + p x^2 + q = (x^2 + 2 x + 5) (x^2 + a x + b) =
=x^4 + (2 + a) x^3 + (5 + b + 2a) x^2 + (5a + 2b) x + 5b




^^^^^^^^ I don't understand this ???

Answer 6

can you explain this?

Answer 7

@hartnn

Answer 8

whats (a+b)(c+d) = ... ?

Answer 9

this is the question and please help me to solve it :-

For x^2+2x+5 to be a factor of x^4+px^2+q, the values of p and q should respectively be :-

a)2,5

b)5,25

c)6,25

d)5,2

Answer 10

ok,
so multiply these out
(x^2 + 2 x + 5) (x^2 + a x + b)

like
x^2 (x^2 + a x + b) +2x (x^2 + a x + b) +5(x^2 + a x + b) =....?

Answer 11

but how can we write x^4 + p x^2 + q into (x^2 + 2 x + 5) (x^2 + a x + b)

Answer 12

x^4+..... is a quartic equation
it can be written as product of 2 quadratic equation

like a cubic equation can be written as
Ax^3+Bx^2+Cx+D = (Ex+F) (Gx^2+Hx+I)

Answer 13

where this comes?
(x^2 + 2 x + 5) (x^2 + a x + b)

Answer 14

see mayank y u dont solve in a simple way

x4 + px2+q = x2 ( x2+ax+b) + 2x ( x2+ax+b ) + 5 ( x2+ax+b )

Answer 15

u tried this way

Answer 16

i feel they u will get answer in this format

Answer 17

you mean why we took specifically x^2+ax+b and not any other quadratic ?

Answer 18

yeah

Answer 19

x4 + px2 +q = x4 + x3 (a +2 ) + x2( b+2a +5 ) + x (2b +5a ) +5b

now compare both side

example x3 in left hand side missing so , we can get
0 = a +2
ie a = -2

and so on u can proceed i belive

Answer 20

i don't understand your method . @vinurules
x4 + px2+q = x2 ( x2+ax+b) + 2x ( x2+ax+b ) + 5 ( x2+ax+b )

?????????????

Answer 21

yes u can break brackets in case of multiplication

Answer 22

u want more illustration ?

Answer 23

yeah

Answer 24

i just opened brackets

Answer 25

but you get why we must take a quadratic there, right ?
to make the degree of polynomial = 4 on right, (bcoz degree of left =4)

and since co-efficient of x^4 is +1
co-efficient of both x^2 term must be 1
and we can take any co-efficients for x and constant, so we took a and b