Question

If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5), (4, 220), and (-2, -20) what would the coefficients a, b, c, and d equal? So confused, I'd greatly appreciate help! Thank you!

Answer 1

Answer:

a = 3

b = 2

c = 0

d = -4

Step-by-step explanation:

Form 4 equations and solve simultaneously

28 = a(2)³ + b(2)² + c(2) + d

28 = 8a + 4b + 2c + d (1)

-5 = -a + b - c + d (2)

220 = 64a + 16b + 4c + d (3)

-20 = -8a + 4b - 2c + d (4)

(1) + (4)

28 = 8a + 4b + 2c + d

-20 = -8a + 4b - 2c + d

8 = 8b + 2d

d = 4 - 4b

Equation (2)

c = -a + b + d + 5

c = -a + b + 4 - 4b+ 5

c = -a - 3b + 9

28 = 8a + 4b + 2c + d (1)

28 = 8a + 4b + 2(-a - 3b + 9) + 4 - 4b

28 = 6a - 6b + 22

6a - 6b = 6

a - b = 1

a = b + 1

220 = 64a + 16b + 4c + d (3)

220 = 64(b + 1) + 16b + 4(-b - 1 - 3b + 9) + 4 - 4b

220 = 60b + 100

60b = 120

b = 2

a = 2 + 1

a = 3

c = -3 - 3(2) + 9

c = 0

d = 4 - 4(2)

d = -4

Answer 2

Step-by-step explanation:

Step 1:  Solve using the first point

(2, 28)

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

Step 2:  Solve using the second point

(-1, -5)

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

Step 3:  Solve using the third point

(4, 220)

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

Step 4:  Solve using the fourth point

(-2, -20)

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

Step 5:  Combine the first and fourth equations

$28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

Step 6:  Solve for c in the second equation

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

Step 7:  Substitute d with the stuff we got in step 5

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

Step 8:  Substitute d and c into the first equation

$28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

Step 9:  Substitute a, b, and c into the third equation

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

Step 10:  Find a using b = 2

$a = b + 1$

$a = (2) + 1$

$a = 3$

Step 11:  Find c using a = 3 and b = 2

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

Step 12:  Find d using b = 2

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer:  $a = 3, b = 2, c = 0,d = -4$