High School

Factor the polynomial \( f(x) \). Then solve the equation \( f(x) = 0 \).

\[ f(x) = x^4 - x^3 - 39x^2 + 109x - 70 \]

The solutions to \( f(x) = 0 \) are:

(Use a comma to separate answers as needed. If a solution occurs more than once, type the solution only once.)

Answer :

Final answer:

The factored form of the polynomial[tex]f(x) = x^4 – x^3 - 39x^2 + 109x - 70[/tex] is [tex](x - 2)(x + 5)(x - 1)(x + 7)[/tex]. The solutions to the equation f(x) = 0 are x = 2, x = -5, x = 1, and x = -7.

Explanation:

To factor the polynomial [tex]f(x) = x^4 – x^3 - 39x^2 + 109x - 70,[/tex] we can use the Rational Root Theorem to find possible rational roots. The Rational Root Theorem states that if a polynomial has a rational root p/q, where p is a factor of the constant term and q is a factor of the leading coefficient, then p/q is a possible root.

In this case, the constant term is -70 and the leading coefficient is 1. The factors of -70 are ±1, ±2, ±5, ±7, ±10, ±14, ±35, and ±70. The factors of 1 are ±1.

By trying out these possible rational roots, we find that x = 2 is a root of the polynomial. Using synthetic division, we can divide the polynomial by (x - 2) to obtain the quotient polynomial.

The quotient polynomial is[tex]x^3 + x^2 - 37x + 35[/tex]. Now, we can continue factoring the quotient polynomial using methods like factoring by grouping or the quadratic formula.

After factoring the polynomial completely, we find that[tex]f(x) = (x - 2)(x + 5)(x - 1)(x + 7).[/tex]

To solve the equation f(x) = 0, we set each factor equal to zero and solve for x:

  1. x - 2 = 0, x = 2
  2. x + 5 = 0, x = -5
  3. x - 1 = 0, x = 1
  4. x + 7 = 0, x = -7

Therefore, the solutions to f(x) = 0 are x = 2, x = -5, x = 1, and x = -7.

Learn more about factoring and solving polynomial equations here:

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