Determine the equation of the quadratic with zeros x = 1 and x = 5 passing through the point (3,-12).

Please help

is the equation the quadratic as per the given conditions.

Step-by-step explanation:

Let the equation of the quadratic be:

Given that it has zeroes at x = 1 and x = 5 i.e. y = 0 at both the values of x.

Also passes through point (3, -12). i.e. when x = 3, y = -12

Putting x = 1, y = 0:

Putting x = 5, y = 0:

Putting x = 3, y = -12:

Equation (2) - Equation (1):

Equation (2) - Equation (3):

Equation (5) - Equation (4):

Putting value of a in equation (4):

Putting a and b in equation (1):

So, the quadratic equation is:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

50- 250 is correct i did that one too

step-by-step explanation: