Respuesta :
The number of real roots found are-
1. (x-4)(x+3)^2(x-1)^3=0 ; 6 real roots.
2. X^2(x^3-1)=0 ; 3 real roots.
3. X(x+3)(x-6)^2=0 ; 4 real roots.
4. 3x(x^3-1)^2=0 ; 3 real roots.
5. (x^3-8)(x^4+1)=0 ; 1 real roots.
Define the term roots of polynomial?
- If a root is not imaginary (defined as a number as in form a + bi, whereby I is a number which is not on any real number line, which is why the name imaginary), it is said to be real.
Consider the given cases-
1. (x-4)(x+3)^2(x-1)^3=0,
Put each equals zero.
x – 4 = 0 or (x + 3)^2 = 0 or (x – 1)^3 = 0,
Thus,
x = 4
x = –3 twice
x = 1 thrice
All are real numbers, thus given polynomial has 6 real roots.
2. x^2(x^3-1)=0
Put each equals zero.
x² = 0
x³ – 1 = x³ – 1³ = (x – 1)(x² + x + 1) = 0
The quadratic expressions has 2 imaginary roots.
Use the discriminant formula to find-
D = b² – 4ac = 1² – 4(1)(1) = 1 – 4 = –3 < 0.
Thus, roots that are not real.
So, the real roots of this polynomial : 0(twice) and 1.
Hence, thus given polynomial has 3 real roots.
3. x(x+3)(x-6)^2=0
Put each equals zero.
x = 0
x = 3
x = 6 twice
Hence, thus given polynomial has 4 real roots.
4. 3x(x^3-1)^2=0
Put each equals zero.
3x = 0 and x = 0
(x^3 – 1)² = [(x – 1)(x² + x + 1)]² = (x – 1)²(x² + x + 1)² = 0,
Here, only two roots are real
The quadratic expression do have unreal roots, thus 1 twice are the real roots.
Hence, thus given polynomial has 3 real roots..
5. (x^3-8)(x^4+1)=0
Put each equals zero.
x³ – 8 = x³ – 2³ = (x – 2)(x² + 2x + 4) = 0
There is only one real root i.e, 2, the quadratic expression do have imaginary roots
Check using the discriminant formula
D = b² – 4ac = 2² – 4(1)(4) = 4 – 16 = – 12 < 0.
x⁴ + 1 = 0 has imaginary roots.
Hence, thus given polynomial has 1 real roots..
To know more about the roots of polynomial, here
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