Find the roots of the given polynomial equation.
2 answers:
9514 1404 393
Answer:
- -1, 1, 2 (multiplicity 2)
- -7, -5, 3
Step-by-step explanation:
Just as you find it easiest to ask someone else to find the roots for you, I find it easiest to use a graphing calculator to find the roots. It shows the roots to be ...
- -1, 1, 2 (multiplicity 2)
- -7, -5, 3
If you're searching for roots by hand, the usual process is to make a guess based on Descartes' Rule of Signs, and the Rational Root Theorem. Then factor out that root using synthetic division or long division to reduce the polynomial degree.
__
<h3>x^4-4x^3+3x^2+4x-4=0</h3>This has 3 sign changes, so 1 or 3 positive real roots. If odd-degree terms change signs, there is one sign change, indicating 1 negative real root. The Rational root theorem tells you the rational roots will be divisors of 4:
±1, ±2, ±4
Both even-degree terms and odd-degree terms have coefficients that sum to zero. This indicates that 1 and -1 are both roots of the equation. Factoring those out gives ...
(x -1)(x +1)(x^2 -4x +4) = 0
The quadratic is a perfect square, so the full factorization is ...
(x +1)(x -1)(x -2)² = 0
Roots are -1, 1, and 2 (multiplicity 2).
__
<h3>x^3+9x^2-x-105=0</h3>The rule of signs tells you there is one positive real root and 0 or 2 negative real roots. The rational roots will be divisors of 105, so will be any of ...
±1, ±3, ±5, ±7, ±15, ±21, ±35, ±105
Various other ways of estimating roots suggest the magnitudes will be less than ∛105 +√1 +9 ≈ 14.7. The sum of coefficients tells you the positive root is greater than 1. Trying x=3 shows that to be the positive real root. Factoring it out gives ...
(x -3)(x² +12x +35) = 0
The quadratic can be further factored, so we have ...
(x -3)(x +5)(x +7) = 0
Roots are -7, -5, and 3.
You might be interested in