Find all real zeros of the polynomial $f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 49 x - 147$ and determine the mutiplicity of each.
A) $x = 7$, multiplicity $2 ; x = - 3$, multiplicity 1
B) $x = 7$, multiplicity $1 ; x = - 7$, multiplicity $1 ; x = - 3$, multiplicity 1
C) $x = - 3$, multiplicity $2 ; x = - 7$, multiplicity 1
D) $x = - 7$, multiplicity $1 ; x = 3$, multiplicity $1 ; x = - 3$, multiplicity 1
E) $\quad x = - 3$, multiplicity 3

Question

Quizplus · Accepted Answer

To find the real zeros of the polynomial $f(x) = x^3 + 3x^2 - 49x - 147$, we can attempt to factor it. One way to start is by using the Rational Root Theorem to test possible rational roots. Given the constant term and the leading coefficient, possible rational roots include factors of 147 (±1, ±3, ±7, ±21, ±49, ±147). Testing these, we find that $x = -3$ is a root because $f(-3) = (-3)^3 + 3(-3)^2 - 49(-3) - 147 = 0$. Dividing the polynomial by $x + 3$ gives us a quotient of $x^2 - 49$, which can be further factored into $(x + 7)(x - 7)$. Therefore, the real zeros are $x = -3$, $x = -7$, and $x = 7$, each with a multiplicity of 1, as each factor appears exactly once in the factored form of the polynomial.

Find All Real Zeros of the Polynomial f(x)=x3+3x2−49x−147f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 49 x - 147f(x)=x3+3x2−49x−147

Find All Real Zeros of the Polynomial $f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 49 x - 147$