Question

Using factor theorem factories x^3+6x^2+11x+6

Answer 1

x^3 + 6x^2+11x+6

factor of 6= 1,2,3,6etc

let

x+ 1=0

x=-1

p(x) = x^3+6x^2 +11x+6

p(-1)= -1^3+6(-1)^2+11(-1)+6

= -1+6-11+6

=-12 +12

= 0

Then ( x+1) is the factor of polynomial x^3+6x^2 +11x+6

By long divisions method

x+1÷x^3+6x^2 +11x+6 =x^2+5x +6

then p(x) =( x+1) (x^2+5x+6)

=( x+1) (x^2 + 2x+3x +6)

=(×+1) [x(x+2]+3(x+2)]

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

These are the factor of x^3+6x^2+11x+6

Answer 2

Answer:

Step-by-step explanation:

x^3 + 6x^2+11x+6

factor of 6= 1,2,3,6etc

let

x+ 1=0

x=-1

p(x) = x^3+6x^2 +11x+6

p(-1)= -1^3+6(-1)^2+11(-1)+6

= -1+6-11+6

=-12 +12

= 0

Then ( x+1) is the factor of polynomial x^3+6x^2 +11x+6

By long divisions method

x+1÷x^3+6x^2 +11x+6 =x^2+5x +6

then p(x) =( x+1) (x^2+5x+6)

=( x+1) (x^2 + 2x+3x +6)

=(×+1) [x(x+2]+3(x+2)]

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