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If 2x2 - 7x-15 is a factor of H(x):- 6x3 - 13x2 - ax - b where a and b are constants.

Find the values of a and b.

Find the zeros of H(X)


1 Answer

1 vote

Answer:

1.) a = 73, b = 60

2.) H(0) = -60

Explanation:

Since equation 2x2 - 7x - 15 is a factor of H(x) = 6x3 - 13x2 - ax - b where a and b are constants.

Find the root of 2x^2 - 7x - 15

2x^2 - 10x + 3x - 15 = 0

2x( x - 5 ) + 3( x - 5 ) = 0

2x + 3 = 0 or x - 5 = 0

2x = -3 or x = 5

X = -3/2 or 5

Substitute the two values into the H(x)

When x = - 3/2

6(-3/2)^3 - 13( -3/2 )^2 - a( -3/2 ) - b = 0

6(-27/8) - 13( 9/4 ) + 3a/2 -b = 0

-81/4 - 117/4 + 3a/2 - b = 0

- 198 + 6a - 4b = 0 .... (1)

When x = 5

6( 5 )^3 - 13( 5 )^2 - a(5) - b = 0

6(125) - 13(25) - 5a - b = 0

750 - 325 - 5a - b = 0

425 - 5a - b = 0 ..... ( 2)

Solve equation 1 and 2 simultaneously

Eliminate b

- 198 + 6a - 4b = 0 × 1

425 - 5a - b = 0 × 4

- 198 + 6a - 4b = 0

1700 - 20a - 4b = 0

- 1898 + 26a = 0

26a = 1898

a = 1898/26

a = 73

Substitute a back into equation 1

- 198 + 6(73) - 4b = 0

-198 + 438 - 4b = 0

4b = 240

b = 240/4

b = 60

2.) To Find the zeros of H(X), substitute 0 for x. that is,

H(0) = 6(0)^3 - 13(0)^2 - 73(0) - 60

H(0) = - 60

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