Question

Find the polynomial of degree 3 such that the coefficient of x2

is three less than the coefficient
of x3
; the coefficient of x is 3 times the coefficient of x2
; and the constant term is two more than
the coefficient of x3
. The sum of the coefficients is -4.

Answer 1

1/5[8x³ –7x² –21x + 18]

Explanation:

Let the polynomial be:

ax³ + bx² + cx + d

Where:

a => is the coefficient of x³

b => is the coefficient of x²

c => is the coefficient of x

d => is the constant term.

From the question given above,

Coefficient of x² (b) = a – 3 ..... (1)

Coefficient of x (c) = 3b ...... (2)

Constant term (d) = 2 + a ..... (3)

Sum of coefficient (a + b + c) = –4

a + b + c = –4

b = a – 3

c = 3b = 3(a – 3)

a + b + c = –4

a + (a – 3) + 3(a – 3) = –4

a + a – 3 + 3a – 9 = –4

Collect like terms

a + a + 3a = –4 + 3 + 9

5a = 8

Divide both side by 5

a = 8/5

Substitute the value of a into equation (1)

b = a – 3

a = 8/5

b = 8/5 – 3

b = (8 – 15)/5

b = –7/5

Substitute the value of b into equation (2)

c = 3b

b = –7/5

c = 3(–7/5)

c = –21/5

Substitute the value of a into equation (3)

d = 2 + a

a = 8/5

d = 2 + 8/5

d = (10 + 8)/5

d = 18/5

SUMMARY:

a = 8/5

b = –7/5

c = –21/5

d = 18/5

Thus, the polynomial:

ax³ + bx² + cx + d

8/5x³ + (–7/5)x² + (–21/5)x + 18/5

8/5x³ –7/5x² –21/5x + 18/5

1/5[8x³ –7x² –21x + 18]