Question

Use the quotient from part (b) to find the remaining roots and solve the equation.

Answer 1

Given that $x^3-2x^2-7x-4=0$

Part a)

The possible rational roots of the polynomial are the factors of the produce of the leading coefficient and the constant term.

Thus, the possible rational roots of the polynomial are the factors of -4.

Therefore, the possible roots of the polynomial are: $\pm1,\ \pm2,\ \pm4$



Part B:

Using the factor theorem, we can get the actual factors.

$f(1)=(1)^3-2(1)^2-7(1)-4=1-2-7-4=-12\neq0 \\ \\ f(-1)=(-1)^3-2(-1)^2-7(-1)-4=-1-2(1)+7-4 \\ =2-2=0$

Thus, x = -1 is an actual root of the polynomial.



Part c):

Given that x = -1 is an actual root of the polynomial. Thus x + 1 is a factor and the other factors can be obtained by dividing the polynomial by x + 1. This gives $x^2-3x-4=(x-4)(x+1)$

Therefore, the roots of $x^3-2x^2-7x-4=0$ are: x = -1 with a multiplicity of 2 and x = 4.