f(x) = 4x^4 + 21x^3 +33x^2 + 147x +35
Find all the zeros using the Rational zero theorem
Polynomials usually higher than 3 degrees is usually solved with synthetic division. This is usually a trial and error method given the factors that can be multiplied together to form either the beginning and ending coefficients 4 and 35 respectively. Both negative and positive values of these factors should be tried.
The important process is to have no remainder left over once we attempt synthetic division. Essentially, the last number we bring down will be zero. If not, we must try another factor and repeat the process.
Using -5 as a potential factor:
-5 |+4 +21 +33 +147 +35 (notice we used the coefficients in front of each degree)
_____-20__-5___-140_-35__ (also note there is a place holder for each degree, from 4 down to 0)
+4 +1 +28 +7 0 (process of synthetic division, bring down the first coefficient (+4), then multiply +4*-5 = -20 and place the -20 on the middle row under the second coefficient (+21). Then add the top and middle rows +21 + -21 = +1. Repeat this process. This is factor since our remainder, the last number on the bottom row, is equal to 0)
So, the first factor or rational zero of this polynomial is -5. This would be represented as (x + 5)
Now we rewrite the equation and see if is now factorable. The numbers on the bottom row are the new coefficients of the partially factored polynomial. Also, the polynomial has decreased by the power of 1 since we factored out an x.
New polynomial: (x+5)(4x3 + 1x2 + 28x + 7)
Since this still doesn't look like it's easily factored. We will have to repeat this process again, with trial and error. The next factor would -1/4.
Through the process of synthetic division, the polynomial looks like this:
(x+5)(x+1/4)(4x2 + 0x + 28) (notice the polynomial decreased another degree and all degree of x from 2 to 0 have placeholders)
From here we can factor the remaining polynomial:
(x+5)(x+1/4)*4(x2 + 7) (since we are finding values of x that make this polynomial equal to zero, we can set the entire polynomial equal to zero. From here, we can divide both sides by 4)
(x+5)(x+1/4)(x2+7) = 0 (finish factoring the x2)
(x+5)(x+1/4)(x+√(-7))(x-√(-7)) = 0
Since there are 4 degrees in the original polynomial, there must be 4 answers or roots for x that make this equation equal to zero.
x = -5, -1/4, -i√7, +i√7 (since the last two roots are not real numbers)
answer: x = -5, -1/4
in addition to Isaac's work, just to show division
4x3 + x2 + 28x + 7
(x+5) _ l 4x4 + 21x3 + 33x2 + 147x + 35
4x4 + 20x3
_ x3 + 33x2
x3 + 5x2
_ 28x2 + 147x
28x2 + 140x
_ 7x + 35
7x + 35
4x3 + x2 + 28x + 7 = (4x3 + x2) + (28x + 7) = x2(4x + 1) + 7(4x + 1) = (4x + 1)(x2 + 7)
4x4 + 21x3 + 33x2 + 147x + 35 = (x + 5)(4x + 1)(x2 + 7)
There are only two zeros in set of real numbers: -1/4 and -5
The overall strategy is to find the rational roots, and then to use factors based on the rational roots to reduce the polynomial degree by division. Once we are working with a polynomial of degree one or two, we can use simple methods to find the remaining roots.
According to the rational zero theorem the possible rational zeroes are all of the possible factors of the constant coefficient, in this case 35, divided by all of the possible factors of the leading coefficient. In this case 4. The factors of 35 are 1, 5, 7, and 35. The factors of 4 are 1, 2, and 4. Try positive and negative combinations using either synthetic division or substitution in the polynomials. By trial and error we find that -1/4 and -5/1 are zeroes. We can write (x+5) and (x+1/4) as factors. To clear fractions, let's use (4x+1) as the second factor instead of using (X+1/4).
Now we can use division and the two factors above to reduce our working polynomial from degree 4 to degree 2. The result is the polynomial X^2 + 7 = 0. This is easily solved as follows:
X^2 = -7; X = +/- i sqrt (7). We have found all four roots
Since all the coefficients are positive, there is no positive solution.
By root theorem, try x = -factor of 35/factor of 4. Since f(-1/4) = f(-5) = 0, using synthetic division,
-5 | 4 21 33 147 35
........-20 -5 -140 -35
....4 1 28 7 | 0
-1/4 | 4 1 28 7
..........-1 0 -7
......4 0 28 | 0
Since 4x^2 + 28 = 0 has no real solution, there are only two zeros. They are x = -5, and x = -1/4.