Answer:
Possible rational roots: ±1,±3,±5,±15.
Please correct me If I am wrong
Step-by-step explanation:
Since all coefficients are integers, we can apply the rational zeros theorem.
The trailing coefficient (coefficient of the constant term) is 15.
Find its factors (with plus and minus): ±1,±3,±5,±15. These are the possible values for p.
he 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,±31,±51,±151.
Simplify and remove duplicates (if any), these are possible rational roots: ±1,±3,±5,±15.
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+6x+15 by x−1.The quotient is x2+4x+10 and the remainder is 25
Check −1: divide x3+3x2+6x+15 by x+1.The quotient is x2+2x+4 and the remainder is 11
Check 3: divide x3+3x2+6x+15 by x−3.The quotient is x2+6x+24 and the remainder is 87
Check −3: divide x3+3x2+6x+15 by x+3.The quotient is x2+6 and the remainder is −3
Check 5: divide x3+3x2+6x+15 by x−5.The quotient is x2+8x+46 and the remainder is 245
Check −5: divide x3+3x2+6x+15 by x+5.The quotient is x2−2x+16 and the remainder is −65
Check 15: divide x3+3x2+6x+15 by x−15.The quotient is x2+18x+276 and the remainder is 4155
Check −15: divide x3+3x2+6x+15 by x+15.The quotient is x2−12x+186 and the remainder is −2775
Possible rational roots: ±1,±3,±5,±15.
Please correct me if I am wrong I did my best its been so long I haven't done rational roots
