Use the rational zeros theorem to find all the real zeros of the polynomial function. Use the zeros to factor f over the real numbers.
f(x)=25x^4+26x^3+126x^2+130x+5
Find the real zeros
x=
Use the real zeros to factor f
f(x)=
only possible rational roots are
x = ±1, ±1/5, ± 1/25
quickly found x=-1 to work
so one factor is x+1
after reducing it to a cubic by synthetic division, it took a bit longer to find x = -1/25 to work
so (25x+1) is another factor
long algebraic divsion gave the last factor as x^2 + 5, which has no real roots.
so real roots are
x = -1 and x = -1/25
just a note:
Things worked out in this case, but x = ±5 were also candidates, since 5/1 has suitable numerator and denominator. For example,
25x^4-100x^3-124x^2-4x-5
has similar coefficients, but has real roots -1 and 5:
(x+1)(x-5)(25x^2+1)
given that f(x) = 9/x-5 and g(x) = 12/x+12 find
To find the real zeros of the polynomial function f(x) = 25x^4 + 26x^3 + 126x^2 + 130x + 5, we can apply the Rational Zeros Theorem.
1. The Rational Zeros Theorem states that if a polynomial has a rational zero (a/b), where "a" is a factor of the constant term and "b" is a factor of the leading coefficient, then the rational zero satisfies the equation f(x) = 0.
2. For the given polynomial f(x) = 25x^4 + 26x^3 + 126x^2 + 130x + 5, the constant term is 5 (a) and the leading coefficient is 25 (b).
3. The factors of 5 are 1 and 5, and the factors of 25 are 1, 5, and 25. Therefore, the possible rational zeros are ±1, ±5, and ±1/5.
4. We can now check each of these possible rational zeros to see if any of them satisfy the equation f(x) = 0. This can be done using synthetic division, plugging in each value and checking if the remainder is zero.
Starting with x = 1:
25 | 25 26 126 130 5
25 51 177 307
-------------------
25 51 177 307 312
Since the remainder is not zero, x = 1 is not a zero.
Now trying x = -1:
25 | 25 26 126 130 5
-25 -1 -125 -5
-------------------
0 25 1 0
Since the remainder is zero, x = -1 is a zero.
Therefore, the real zeros of the polynomial function f(x) = 25x^4 + 26x^3 + 126x^2 + 130x + 5 are x = -1.
To factor f(x) over the real numbers, we can divide f(x) by (x - (-1)). Using synthetic division:
-1 | 25 26 126 130 5
-25 -1 -125 -5
-------------------
25 1 125 5 0
The resulting quotient is 25x^3 + 1x^2 + 125x + 5, and the factorized form of f(x) is:
f(x) = (x + 1)(25x^3 + 1x^2 + 125x + 5)