Question

factorize: 4x^3+20x^2+33x+18

Answer 1

4x^3 +20x^2 +33x + 18
=> 4x^3 +8x^2 +12x^2 + 24x + 9x + 18
=> 4x^2 (x+2) + 12x(x+2) + 9(x+2)
=> (x+2)(4x^2 +12x +9)
=> (x+2)(4x^2 +6x + 6x +9)
=> (x+2)[2x(2x+3) + 3(2x+3)]
=> (x+2)(2x+3)(2x+3)

Hope it help you

Please mark as brainliest if you liked the solution

Answer 2

According to factor theorem  a polynomial g(x) is a factor of (x-k)  if and only if f(k)=0 (that is k is the root). We are given the polynomial 4x³+20x²+33x+18. Let us say that p(x)=4x³+20x²+33x+18. Then by factor theorem, (x-k) is a factor if p(k)=0. So we need to check weather the factors of 18 are giving us p(k) =0. The factors of 18 are +-1,+-2,+-3,+-6,+-18. Let us substitute these numbers into our polynomial. By trail and error we find that p(-2) gives us remainder 0. Hence, p(-2) = (2-k)(q-r) =0  2-k=0 k=2 .Therefore x+2 is a factor of p(x). Now let us divide p(x) by x+2 . Give it a try! after dividing we get p(x)=(x+2)(4x²+20x+9) . Now we have to split the middle term such that their sum gives 20x and their product gives us 36x². we get 
p(x)=(x+2)(4x²+2x+18x+9)
⇒(x+2)(2x(2x+1)+9(2x+1)
⇒p(x)=(x+2)(2x+1)(2x+9)

I hope that helps!