Write the function as a product of linear and irreducible quadratic factors, all with real coefficients.
-$$f ( x ) = x ^ { 3 } - 6 x ^ { 2 } - 3 x - 28$$
A) $( x - 7 ) \left( x ^ { 2 } + x + 4 \right)$
B) $( x + 7 ) \left( x ^ { 2 } + x + 4 \right)$
C) $( x + 7 ) \left( x ^ { 2 } + x - 4 \right)$
D) $( x - 7 ) \left( x ^ { 2 } + x - 4 \right)$

Question

Quizplus · Accepted Answer

To factor the given cubic polynomial, we first look for rational roots using the Rational Root Theorem. By trying possible factors of the constant term (-28), we find that $x = 7$ is a root, since $f(7) = 7^3 - 6(7)^2 - 3(7) - 28 = 0$. Dividing the polynomial by $x - 7$ gives the quadratic factor $x^2 + x + 4$, which cannot be factored further over the reals because its discriminant $b^2 - 4ac = 1^2 - 4(1)(4) = 1 - 16 = -15$ is negative. Thus, the factorization is $(x - 7)(x^2 + x + 4)$.

Write the Function as a Product of Linear and Irreducible f(x)=x3−6x2−3x−28f ( x ) = x ^ { 3 } - 6 x ^ { 2 } - 3 x - 28f(x)=x3−6x2−3x−28

Write the Function as a Product of Linear and Irreducible $f ( x ) = x ^ { 3 } - 6 x ^ { 2 } - 3 x - 28$