Question

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Answer 1

Answer:

Possible rational roots: ±1,±5.

Actual rational roots: −1,5.

Step-by-step explanation:

The trailing coefficient (coefficient of the constant term) is −5.

Find its factors (with plus and minus): ±1,±5. These are the possible values for p.

The leading coefficient (coefficient of the term with the highest degree) is 1.

Find its factors (with plus and minus): ±1. These are the possible values for q.

Find all possible values of pq: ±11,±51.

Simplify and remove duplicates (if any), these are possible rational roots: ±1,±5.

Next, check the possible roots: if a is a root of the polynomial P(x), the remainder from the division of P(x) by x−a should equal 0.

Check 1: divide x3−3x2−9x−5 by x−1.The quotient is x2−2x−11 and the remainder is −16  

Check −1: divide x3−3x2−9x−5 by x+1.The quotient is x2−4x−5 and the remainder is 0  

So, −1 is a root.

Check 5: divide x3−3x2−9x−5 by x−5.The quotient is x2+2x+1 and the remainder is 0  

So, 5 is a root.

Check −5: divide x3−3x2−9x−5 by x+5.The quotient is x2−8x+31 and the remainder is −160  

Answer 2

1. Find all of the possible roots by using ±p/±q p= the constant
q= lead coefficient of highest term which is x^3 which has a leading coefficient of 1.
2. P=±1,±5, Q=±1. You would divide the factors of the constant by the factors of the leading coefficient.
3. Possible roots for this equation. ±1,±5.
4. Factor the equation.

x³-3x²−9x−5

x³+x²−4x²−4x-5x-5

x²(x+1)-4x(x+1)-5(x+1)

(x+1)(x²-4x-5)

(x+1)(x-5)(x+1)

5. Set all of the factors to zero and solve for x to get your rational roots to the equation.

x+1-1=0-1
x=-1

x+1-1=0-1
x=-1

x-5+5=0+5
x=5

6. Final Solution:

Possible Rational Roots: ±1, ±5
Rational Zeros: (5,-1 multiplicity of 2)
Answer Choice: D