(x+3)^5 + (x+3)^2

Question

factorize completely

Answer 1

f(x) = (x+3)^5 + (x+3)^2

= (x+3)^2 ((x+3)^3 + 1)

= (x+3)^2 ((x+3) + 1)((x+3)^2 - (x+3) + 1) =

= (x+3)^2 (x+4) (x^2 + 6x + 9 - x - 2)

= (x+3)^2 (x+4) (x^2 + 5x + 7)

If x^2 + 5x + 7 = 0, then

x = (-b ± √(b^2 - 4ac))/(2a)

x = (-5 ± √(5^2 - 4(1)(7)))/(2*1)

x = (-5 ± √(25 - 28))/2

x = (-5 ± √(-3))/2

x = (-5 ± i√(3))/2 = -5/2 ± i(√(3))/2

x^2 + 5x + 7 = (x - (-5/2 + i(√(3))/2))(x - (-5/2 - i(√(3))/2))

x^2 + 5x + 7 = (x + 5/2 - i(√(3))/2))(x + 5/2 + i(√(3))/2))

f(x) = (x + 3)^2 (x + 4) (x + 5/2 - i(√(3))/2)) (x + 5/2 + i(√(3))/2))

 

Answer 2

Hi Hammad,

 

(x+3)^5+(x+3)^2

(x+3)^2 {(x+3)^3+1)}............(Common factor (x+3)^2)

(x+3)^2 {x^3+3x^2*3+3x*3^2+3^3+1}..............( using formula,(x+y)^3=x^3+3x^2y+3xy^2+y^3))

By simplifying,

(x+3)^2 {x^3+9x^2+27x+27+1}

(x+3)^2 {x^3+9x^2+27x+28}

(x+3)^2 {x^3+4x^2+5x^2+20x+7x+28}

(x+3)^2 {x^2(x+4)+5x(x+4)+7(x+4)}

(x+3)^2 (x+4){x^2+5x+7}............1

 

For factors of x^2+5x+7

 

using formula,if eq ax^2+bx+c=0, x=-b+sq root of (b^2-4ac)/2a or x=-b-sqroot of(b^2-4ac)/2a

 

substituting a=1,b=5,c=7,we get,

x=-5+sq root of(25-4*1*7)/2*1=-5+sq root of(25-28)/2=-5+sq root of -3/2=-5+(sq root of3)*i/2 

(As sq root of -1=i, complex number)

similarly x=-5-(sq root of 3)*i/2 

so factors are {x-(-5+sq root of 3*i/2)}{x-(-5-sq root of 3*i/2)}

plug in 1,we get,

Ans: (x+3)^2(x+4){x-(-5+sq root of 3*i/2)}{x-(-5-sq root of 3*i/2)}

 

Meena from Strongsville.