Obtain all the zeroes of the polynomial of p(x) if sum of all the polynomials are given : (1). P (x) = x^4-8x^3+23x^2-28x+12 if 2 zeroes are 1 & 2.(2). p (x)=96x^4-128x^3+62x^2-13x+1 & 2 zeroes are 1/2 & 1/4
#1
checking for the given roots
P(1) = 1 - 8 + 23 - 28 + 12 = 0
yup, x-1 is a factor
P(2) = 16 - 64 + 92 - 56 + 12 = 0
yup, x-2 is a factor
now divided x^4-8x^3+23x^2-28x+12 by (x-1) to get a cubic, then divide that cubic by (x-2),
leaving you with a quadratic expression
set that equal to zero and solve to get two more solutions.
Hint: you will get two more integers, one of them will be the same as one of those given.
Do #2 the same way
hint: the factors you need for the divisions are
(2x-1) and (4x-1)
hint#2: all zeroes will be simple fractions
To obtain all the zeros of a polynomial, we can use the factor theorem and synthetic division.
For the first polynomial, P(x) = x^4-8x^3+23x^2-28x+12, we are given that two of the zeros are 1 and 2.
To find the remaining zeros, we can use synthetic division.
Step 1: Perform synthetic division by dividing P(x) by (x - 1) to obtain a new polynomial.
Divide P(x) by (x - 1):
---------
1 | 1 -8 23 -28 12
| - 1 -7 16
-----------
1 -9 16 -12 28
The result after the first division is a new polynomial, which is:
x^3 - 9x^2 + 16x - 12
Step 2: Perform synthetic division again using the new polynomial and divide by (x - 2).
Divide (x^3 - 9x^2 + 16x - 12) by (x - 2):
---------
2 | 1 -9 16 -12
| 2 -14 4
-----------
1 -7 2 -8
The result after the second division is a new quadratic polynomial, which is:
x^2 - 7x + 2
Step 3: Solve the quadratic polynomial, x^2 - 7x + 2, for its zeros. You can use factoring or the quadratic formula for this step.
Using factoring, we can rewrite x^2 - 7x + 2 as (x - 1)(x - 2). Therefore, the remaining zeros are 1 and 2.
So, the zeros of the polynomial P(x) = x^4-8x^3+23x^2-28x+12 are 1, 2, 1, and 2.
Now, let's move on to the second polynomial, p(x) = 96x^4-128x^3+62x^2-13x+1. We are given that two zeros are 1/2 and 1/4.
Again, we can use synthetic division to find the remaining zeros.
Step 1: Perform synthetic division by dividing p(x) by (2x - 1) to obtain a new polynomial.
Divide p(x) by (2x - 1):
---------
1/2 | 96 -128 62 -13 1
| 48 -40 11 -1
-----------
96 8 22 -2 0
The result after the first division is a new polynomial, which is:
96x^3 + 8x^2 + 22x - 2
Step 2: Perform synthetic division again using the new polynomial and divide by (4x - 1).
Divide (96x^3 + 8x^2 + 22x - 2) by (4x - 1):
---------
1/4 | 96 8 22 -2
| 24 8 8
-----------
96 32 30 6
The result after the second division is a new polynomial, which is:
96x^2 + 32x + 30x + 6
Step 3: Solve the quadratic polynomial, 96x^2 + 62x + 6, for its zeros.
Using the quadratic formula, we find:
x = (-b +/- sqrt(b^2 - 4ac)) / 2a
For 96x^2 + 62x + 6, a=96, b=62, and c=6.
x = (-62 +/- sqrt(62^2 - 4*96*6)) / (2*96)
x = (-62 +/- sqrt(3844 - 2304)) / 192
x = (-62 +/- sqrt(1540)) / 192
x = (-62 +/- 39.274)/192
So the remaining zeros are approximately 0.148 and -0.447.
Therefore, the zeros of the polynomial p(x) = 96x^4-128x^3+62x^2-13x+1 are 1/2, 1/4, 0.148, and -0.447.