x5−8x4+28x3−56x2+64x−32=0
tried: ±1 , ±2 and found
x = 2 to be a triple root.
successive synthetic divisions shows
x^5−8x^4+28x^3−56x^2+64x−32=0
(notice how we show exponents)
x=2 to be triple root
so
x^5−8x^4+28x^3−56x^2+64x−32=0
(x-2)^3 (x^2 - 2x + 4) = 0
our remaining quadratic has no real roots
x = (2 ± √-12)/2
= 1 ± √-3
= 1 ± i√3
so x = 2 , 1 ± i√3
verification:
http://www.wolframalpha.com/input/?i=x%5E5−8x%5E4%2B28x%5E3−56x%5E2%2B64x−32%3D0
To solve the equation x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = 0, you can use the rational root theorem along with synthetic division.
Step 1: Find the possible rational roots
The rational root theorem states that if a polynomial equation has a rational root, it must be in the form of p/q, where p is a factor of the constant term (-32 in this case) and q is a factor of the leading coefficient (1 in this case). The factors of -32 include ±1, ±2, ±4, ±8, ±16, and ±32, and the factors of 1 are ±1. So the possible rational roots are:
±1, ±2, ±4, ±8, ±16, and ±32.
Step 2: Try the possible roots using synthetic division
Start by trying one of the possible roots, such as x = 1.
Divide the equation by (x - 1) using synthetic division:
1 | 1 -8 28 -56 64 -32
| 1 -7 21 -35 29
|________________________________
1 -7 21 -35 29 -3
The result of synthetic division is 1x^4 - 7x^3 + 21x^2 - 35x + 29 - 3/(x - 1).
The remainder is -3, so x = 1 is not a root.
Step 3: Continue trying other possible roots
Repeat the process with the remaining possible rational roots until a root is found.
Let's try another possible root, x = 2.
2 | 1 -8 28 -56 64 -32
| 2 -12 32 -48 32
|________________________________
1 -6 20 -24 16 0
The result of synthetic division is 1x^3 - 6x^2 + 20x - 24. The remainder is 0, so x = 2 is a root.
Step 4: Factor the remaining polynomial
Now that we have found one root, x = 2, we can factor the remaining polynomial, which is now 1x^3 - 6x^2 + 20x - 24.
1x^3 - 6x^2 + 20x - 24 = (x - 2)(1x^2 + 4x + 12)
The factors are (x - 2)(x^2 + 4x + 12).
So the complete factored form of the equation x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = 0 is (x - 2)(x^2 + 4x + 12) = 0.
Step 5: Solve for the remaining roots
To solve for the remaining roots, set each factor equal to zero and solve for x:
For x - 2 = 0, x = 2 (already found)
For x^2 + 4x + 12 = 0, you can use the quadratic formula:
x = (-4 ± √(4^2 - 4(1)(12))) / (2(1))
x = (-4 ± √(-32)) / 2
Since the discriminant (-32) is negative, there are no real roots for x^2 + 4x + 12 = 0.
Therefore, the only real root of the equation x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = 0 is x = 2.
The given equation is a polynomial equation of degree 5. In order to find the values of x that satisfy this equation, we can use the techniques of factoring, synthetic division, or solving numerically using a graphing calculator or computer software.
To solve this equation, we will use the factoring technique.
First, we can observe that all the coefficients in the equation are multiples of 2, so we can divide the entire equation by 2 to simplify it:
(x^5 - 4x^4 + 14x^3 - 28x^2 + 32x - 16) = 0
Next, we look for any common factors among the terms. In this case, we can factor out a common factor of 2:
2(x^5 - 4x^4 + 14x^3 - 28x^2 + 32x - 16) = 0
Now let's try to factor the remaining expression inside the parentheses. Unfortunately, this polynomial does not seem to have any easily identifiable factors. In such cases, we can use synthetic division to check if it has any rational roots.
One way to find possible rational roots is by using the Rational Root Theorem. According to the Rational Root Theorem, the possible rational roots of a polynomial with integer coefficients are all the possible ratios of factors of the constant term (in this case, -16) over factors of the leading coefficient (in this case, 1).
So, the possible rational roots of the given polynomial are the ratios of factors of 16 (the absolute value of -16) over factors of 1. The factors of 16 are ±1, ±2, ±4, ±8, and ±16, while the factors of 1 are ±1. Therefore, the possible rational roots are ±1, ±2, ±4, ±8, and ±16.
Now, we can use synthetic division to test these possible rational roots. Let's start with x = 1:
1 | 1 - 4 14 -28 32 -16
| 1 -3 11 -17 15
-----------------------
1 -3 11 -17 -1
The remainder is -1, which means x = 1 is not a root of the polynomial.
We can repeat this process for the other possible rational roots until we find a root or exhaust all possibilities.
By testing all the possible rational roots, we find that the polynomial has a rational root of x = 2.
Using synthetic division again, we divide the polynomial by (x - 2):
2 | 1 - 4 14 -28 32 -16
| 2 -4 20 -16 32
-------------------------
1 -2 10 -8 16
The result is a cubic equation: x^3 - 2x^2 + 10x - 8 = 0.
Now, we can continue factoring the cubic equation. As it does not seem to have any easily identifiable factors, we can use numerical methods or graphing calculators/software to find the remaining roots.
By using numerical methods or graphing calculators/software, we find that the cubic equation has two more roots: x ≈ 0.5829 and x ≈ 1.4171.
Therefore, the original equation x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = 0 has three real roots: x = 2, x ≈ 0.5829, and x ≈ 1.4171.