Question

Find the roots of the given polynomial equation.

Answer 1

Answer:

Consider x^ {2}-4x+3. Factor the expression by grouping. First, the expression needs to be rewritten as x^ {2}+ax+bx+3. To find a and b, set up a system to be solved. a=-3 b=-1. Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system so

Step-by-step explanation:

Answer 2

9514 1404 393

Answer:

1. -1, 1, 2 (multiplicity 2)
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, 1, 2 (multiplicity 2)
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.

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x^4-4x^3+3x^2+4x-4=0

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).

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x^3+9x^2-x-105=0​

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.