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

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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 $f(-3) = 0$. Factoring $x + 3$ out of $f(x)$ gives us:$$f(x) = (x + 3)(x^2 - 49) = (x + 3)(x + 7)(x - 7)$$This factorization shows that the real zeros are $x = -3$, $x = -7$, and $x = 7$, each with multiplicity 1, which matches option B.

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 $f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 49 x - 147$