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Sagot :
Answer and Step-by-step explanation:
A root is considered a real root if it is not an imaginary number (a number in the form of a + bi, where i is a number that is not on the real number line, hence the name: imaginary). Let's inspect the given polynomials.
1. (x-4)(x+3)^2(x-1)^3=0, the equation implies
x – 4 = 0 or (x + 3)^2 = 0 or (x – 1)^3 = 0, this gives:
x = 4
x = –3 twice
x = 1 thrice
They are all real numbers, therefore it has 6 real roots.
2. x^2(x^3-1)=0
x² = 0
x³ – 1 = x³ – 1³ = (x – 1)(x² + x + 1) = 0
The quadratic expressions gives 2 imaginary roots. To check this we can use the determinant formula.
D = b² – 4ac = 1² – 4(1)(1) = 1 – 4 = –3 < 0. This gives roots that are not real.
Therefore the real roots of this polynomial are: 0(twice) and 1. Hence, it has 3 real roots.
3. x(x+3)(x-6)^2=0
x = 0
x = 3
x = 6 twice
Thus, it has 4 real roots.
4. 3x(x^3-1)^2=0
3x = 0 and this gives x = 0
(x^3 – 1)² = [(x – 1)(x² + x + 1)]² = (x – 1)²(x² + x + 1)² = 0, only two roots are real here (the quadratic expression has unreal roots, therefore 1 twice are the real roots).
Therefore, this has 3 real roots.
5. (x^3-8)(x^4+1)=0
x³ – 8 = x³ – 2³ = (x – 2)(x² + 2x + 4) = 0
The only real root here is 2, the quadratic expression had imaginary roots and can be checked using the determinant formula
D = b² – 4ac = 2² – 4(1)(4) = 4 – 16 = – 12 < 0.
x⁴ + 1 = 0 has imaginary roots. Therefore, this polynomial has only one real root.
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