Question

x^3 + 6x^2 + 12x + 16..........

plzzz solve it fast

Answer 1

First we find out the factor of p(x), that is, the equation which is equal to -4.
Now we write the factor 3 times in the following way, and then put the coefficients by comparing with the equation.
   x^2(x+4)     +2x (x+4)     +4 (x+4)
Now we combine them.
Then, (x+4)(x^2+2x+4)=0
It does not have any real roots. But if we take the root, answer is -28. Therefore x= (-2 + or - root of -28)/4. So this is the answer.
Hope it helps you.

Answer 2

let p(x) =x^3+6x^2+12x+16
=x^3+4x^2+2x^2+8x+4x+16
=x^2(x+4)+2x(x+4)+4(x+4)
=(x+4)(x^2+2x+4)
to find factors of p(x)
take p(x)=0
x+4=0 or x^2+2x+4=0
x=-4
ii) compare 2x^2+2x+4=0 with ax^2+bx+c=0
a=2,b=2,c=4
discriminant = b^2-4ac
=2^2-4*2*4
=4-32
=-28
Therefore
x= [-b+ or - root b^2-4ac]/2a
=[-2+ or - root(-28)]/4