Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. write the polynomial in standard form. 5, -3, and -1 + 3i f(x) = x4 + 12.5x2 - 50x - 150 f(x) = x4 - 4x3 + 15x2 + 25x + 150 f(x) = x4 - 4x3 - 15x2 - 25x - 150 f(x) = x4 - 9x2 - 50x - 150 question 13(multiple choice worth 5 points) use the rational zeros theorem to write a list of all potential rational zeros. f(x) = x3 - 7x2 + 9x - 24 ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 ±1, ±2, ±3, ±4, ±24 ±1, ±one divided by two, ±2, ±3, ±4, ±6, ±8, ±12, ±24 ±1, ±2, ±3, ±4, ±6, ±12, ±24 question 14(multiple choice worth 5 points) perform the requested operation or operations. f(x) = 7x + 6, g(x) = 4x2 find (f + g)(x). 7x + 6 + 4x2 28x3 + 24x 7x + 6 - 4x2 seven x plus six divided by four x squared. will give great feedbback and will give best answer brainliestasap

Q1 - D. f(x) =

Q13 - A. ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.

Q14 - A.

Step-by-step explanation:

Question 1:

We know that rational roots always occurs in pairs. So, the zeros of the function will be 5, -3, -1+3i, -1-3i

So, the factored form is

i.e.

i.e.

i.e.

Hence, the polynomial function is .

Question 13:  

Rational Zeros Theorem states that 'If p(x) is a polynomial with integer coefficients and if is a zero of p(x) = 0. Then, p is a factor of the constant term of p(x) and q is a factor of the leading coefficient of p(x)'.

Let, is a zero of . Then, p is a factor of -24 and q is a factor of 1.

Thus, possible values of p = ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 and q = ±1

This gives, possible values of are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.

Question 14:

We have, f(x) = 7x + 6 and g(x) =

Then, (f+g)(x) = f(x) + g(x) =  7x + 6 +

So, (f+g)(x) =

system of linear equation

step-by-step explanation:

as given in the data

a= 20 , b= -12

let us suppose that it is a straight line equation

now we know the formula for the system of linear equation

b= ma + c )

here m is slope of a straight line which is equal to formula

m=   (ii)

and here c is a constant

now putting the value of a and b in equation (ii)

m=

m=

putting the value of a,b and m in equation (i) gives us

-12 = + c

-12=-12 + c

adding 12 to both sides

-12 +12 = -12 +12 +c

0 =c

or c=0

as the value of the constant is zero which gives us that it is a linear equation and this order pair satisfies the system of linear equations.