If f(x)=x^3+11x^2+31x+21 and x+1 is a factor of f(x)f(x), then find all of the zeros of f(x)f(x) algebraically.

9514 1404 393

  {-1, -3, -7}

Step-by-step explanation:

Dividing the given polynomial by (x+1) yields a quotient of x^2 +10x +21. That factors as (x+3)(x+7), so the three zeros are ...

  x ∈ {-1, -3, -7}

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The factors of x^2 +10x +21 are found by looking for factors of 21 that have a sum of 10. Those factors are 3 and 7, so the complete factorization of f(x) is ...

  f(x) = (x +1)(x +3)(x +7)

The zeros are the values of x that make these factors be zero. In each case, that x-value is the opposite of the constant in the binomial term.

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof

2 pecks

step-by-step explanation: