Question

g(x)=4x^3+qx^2+rx+3. 1)if g(x) is divided by (x-1) it leaves a remainder of -12. 2)(x-3) is a factor of g(x) Use the two facts given above to set up two equations with two unknowns. Hence solve the equations simultaneously to determine the values of q and r

Answer 1

q = - 9 and r = - 10

Explanation:

since g(x) is divided by (x - 1) with remainder - 12, then

g(1) = 4(1)³ + q(1)² + r + 3 = - 12, that is

4 + q + r + 3 = - 12

q + r + 7 = - 12 ( subtract 7 from both sides )

q + r = - 19 → (1)

Since (x - 3) is a factor of g(x), then

g(3) = 4(3)³ + q(3)² + 3r + 3 = 0, that is

108 + 9q + 3r + 3 = 0

9q + 3r + 111 = 0 ( subtract 111 from both sides )

9q + 3r = - 111 → (2)

The 2 equations to be solved simultaneously are (1) and (2)

Multiply (1) by - 3

- 3q - 3r = 57 → (3)

Add (2) and (3) term by term to eliminate r

6q = - 54 ( divide both sides by 6 )

q = - 9

Substitute q = - 9 into (1)

- 9 + r = - 19 ( add 9 to both sides )

r = - 10

Answer 2

Explanation:

When g(x) is divide by (x -1), the remainder = -12

g(1) = -12

4 + q + r + 3 = -12

q+r + 7 = -12

Subtract 7 form both sides.

q + r +7 -7 = -12 - 7

q + r = -19 --------------------(i)

(x -3) is a factor of g(x). So, the remainder = 0

g(3) = 0

4*(3)³ + q*(3)² + r*3 + 3 = 0

4*27 + 9q + 3r + 3 = 0

108 + 9q + 3r +3 = 0

9q +3r + 111 = 0

Subtract 111 from both sides

9q + 3r = -111 -----------------(ii)

Multiply equation (i) by (-3).

(i)*(-3) -3q - 3r = +57

(ii0 9q + 3r = -111 {Now add & r will be eliminated}

6q = -54

q = -54/6

q = -9

Plugin the value of q in equation (i)

-9 + r = -19

Add 9 to both the sides

r = -19 + 9

r = -10