SOLUTION: Use the Rational Root Theorem to find all the roots of each equation.
4. x'3 + 9x'2 + 19x – 4 = 0
5. 2x'3 – x'2 + 10x – 5 = 0
6. Two roots of a polynomial eq
Algebra ->
Rational-functions
-> SOLUTION: Use the Rational Root Theorem to find all the roots of each equation.
4. x'3 + 9x'2 + 19x – 4 = 0
5. 2x'3 – x'2 + 10x – 5 = 0
6. Two roots of a polynomial eq
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Question 174370: Use the Rational Root Theorem to find all the roots of each equation.
4. x'3 + 9x'2 + 19x – 4 = 0
5. 2x'3 – x'2 + 10x – 5 = 0
6. Two roots of a polynomial equation with real coefficients are 2 + 3i and {7} ({} means square root) . Find two additional roots. Then find the degree of the polynomial.
You can put this solution on YOUR website! 4. x^3 + 9x^2 + 19x – 4 = 0
I graphed it and found a zero at x=-4
Then using synthetic division you get:
-4)....1....9....19....-4
........1....5....-1...|..0
Quotient: x^2+5x-1
Use the quadratic formula to find the remaining zeroes:
x = [-5 +- sqrt(25 -4*1*-1)]/2
x = [-5 +- sqrt(29)]/2
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5. 2x^3 – x^2 + 10x – 5 = 0
I graphed it and found a zero at x=1/2
Then used synthetic division to get:
1/2)....2....-1....10....-5
.........2....0.....10...|..0
Quotient: 2x^2 + 10
Therefore the two other zeroes come from
2x^2+10 = 0
x^2 + 5 = 0
x^2 = -5
x = +/- isqrt(5)
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6. Two roots of a polynomial equation with real coefficients are 2 + 3i and {7} ({} means square root) . Find two additional roots. Then find the degree of the polynomial.
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2-3i must be a zero if the coefficients are Real Numbers.
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Comment: Your problem seems to imply that sqrt(7) must be paired with -sqrt(7)
But that is not true.
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The answer to the problem you have posted is:
The degree of the polynomial is three.
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Comment: If that 2nd zero you listed is really isqrt(7) then -isqrt(7)
is a zero and the degree of the polynomial is four.
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Cheers,
Stan H.
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