A polynomial function of least degree with integral coefficients that has the
given zeros 
Given
Given zeros are 3i, -1 and 0
complex zeros occurs in pairs. 3i is one of the zero
-3i is the other zero
So zeros are 3i, -3i, 0 and -1
Now we write the zeros in factor form
If 'a' is a zero then (x-a) is a factor
the factor form of given zeros
Now we multiply it to get the polynomial
polynomial function of least degree with integral coefficients that has the
given zeros
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Answer:
x = - 
, x = 
Step-by-step explanation:
to find the points of intersection equate the 2 equations , that is
7x - 15 = 10 + 12x - 6x² ( subtract 10 + 12x - 6x² from both sides )
6x² - 5x - 25 = 0 ← factor the quadratic on left side
consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term
product = 6 × - 25 = - 150 and sum = - 5
the factors are - 15 and + 10
use these factors to split the x- term
6x² - 15x + 10x - 25 = 0 ( factor the first/second and third/fourth terms )
3x(2x - 5) + 5(2x - 5) = 0 ← factor out (2x - 5) from each term
(2x - 5)(3x + 5) = 0
equate each factor to zero and solve for x
3x + 5 = 0 ⇒ 3x = - 5 ⇒ x = -
2x - 5 = 0 ⇒ 2x = 5 ⇒ x =
Not sure but I think it's (3)
Answer:
multiply 10 with the value given for x and add 2 to the result
Answer:
b=2A/h-a
Step-by-step explanation:
A=1/2h(a+b)
a+b=A/(1/2h)
a+b=A/(h/2)
a+b=(A/1)(2/h)
a+b=2A/h
b=2A/h-a
