Answer: x = -3/4 can not be a rational zero of the polynomial.
Step-by-step explanation:
We have the polynomial:
6x^5 + ax^3 -bx -12 = 0.
The theorem says that:
If P(x) is a polynomial with integer coefficients, and p/q is a zero of P(x) then p is a factor of the constant term (in this case the constant term is -12) and q is a factor of the leading coefficient (in this case the leading coefficient is 6.).
The factors of -12 (different than itself) are (independent of the sign).
1, 2, 3, 4 and 6.
So p can be: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6.
The factors of 6 are:
1, 2 and 3, so q can be 1, -1, 2, -2, 3, -3.
Then the option that can not be a zero of the polynomial is
x = -3/4
because the number in the denominator must be a factor of the leading coefficient, and 4 is not a factor of six.
Answer:
2.15 FT
Step-by-step explanation:
Answer:
x = 48
Step-by-step explanation:
3x + 9 = 153
subtract 9 from both sides -9 -9
3x = 144
divide both sides by 3 3x/3 = 144/3
2x + (x + 9) = 153
2(48) + (48 + 9) = 153
96 + 57 = 153
153 = 153
Answer:
Step-by-step explanation:
<u>See steps below, solving for x:</u>
- x^(p/q) = a
- log x^(p/q) = log a
- p/q log x = log a
- log x = q/p log a
- log x = log a^(q/p)
- x = a^(q/p)