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1 answer:
4. A graph shows the solutions to be
x = -3/2, -1, 0
The root at -1 has multiplicity 2. That is, the factoring is
x(2x+3)(x +1)² = 0
5. A graph shows the real roots to be
x = -1, 2
Dividing these from the polynomial gives the quadratic
x^2 +3
which has roots
x = ±i√3
The factorization is
(x +1)(x -2)(x² +3) = 0
and the solutions are
x = -1, 2, -(√3)i, (√3)i
x = -3/2, -1, 0
The root at -1 has multiplicity 2. That is, the factoring is
x(2x+3)(x +1)² = 0
5. A graph shows the real roots to be
x = -1, 2
Dividing these from the polynomial gives the quadratic
x^2 +3
which has roots
x = ±i√3
The factorization is
(x +1)(x -2)(x² +3) = 0
and the solutions are
x = -1, 2, -(√3)i, (√3)i
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