# Given that (x-2) and (x+3) are factors of 2x4-ax3-10x2+bx-54, find a and b.HELP!

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Let f(x) = 2x4-ax3-10x2+bx-54

since x-2 and x+3 are factors of f(x), we can write

f(x) = 2x4-ax3-10x2+bx-54 = (x-2)(x-3) Q(x)................(1), where Q(x) is the quotient.

Putting x=2in eq(1), we get: 2*2^4-a*2^4-10*2^4+b*2-54 = 0 as the factor x-2 is 0 when x=2. Or 32-8a-40+2b-54 = 0. Or -8a+2b = 62. Dividing by 2, we get:

-4a+b = 31.........(2)

Putting x=-3 in eq(1) , we get: 2*(-3)^4-a(-3)^3-10(-3)^2+b(-3)-54 = 0, as the factor x+3 =3 when x=-3. So, 162+27a-90-3b-54=0. Or 27a-3b = 18. Dividing by 3 we get:

9a-b = 6..........(3).

Eq(2)+Eq(3) gives: 5a = 37 or **a **= 37/5 =**7.2. **Substituting this value of a in (2), we get: 9(7.2)+b = 31 Or b = 31-9*7.2. So **b**= 31-64.8 = **-33.8.**

The given expression is:

2x^4 - ax^3 -10x^2 + bx - 54

As (x - 2) is a factor of the given expression:

For x = 2 the value of factor (x - 2) = (2 - 2) = 0

As one factor is 0 the value of the complete expression is also zero. Therefor substituting value x = 2 in the given expression and equating it to 0 we get:

2(2^4) - a(2^3) -10(2^2) + b*2 - 54 =54

32 - 8a - 40 + 2b - 54 = 0

- 8a + 2b = 62 ... (1)

Similarly:

As (x + 3) is a factor of the given expression:

For x = -3 the value of factor (x + 3) = (- 3 + 3) = 0

As one factor is 0 the value of the complete expression is also 0. Therefore substituting value x = - 3 in the given expression and equating it to 0 we get:

2(3^4) - a(3^3) -10(3^2) + b*(-3) - 54 =54

162 + 27a - 90 - 3b - 54 = 0

27a - 3b = - 18

9a - b = - 6 ... (2)

Multiplying equation 2 by 2 we get:

a + 2b = 62

18a - 2b = - 12 ... (3)

Adding equations (1) and (3) we get:

- 8a + 18a + 2b - 2b = 62 - 12

10a = 50

Therefore:

a = 50/10 = 5

Substituting this value of a in equation (2) we get:

9*5 - b = - 6

45 - b = - 6

b = - 6 - 45 = - 51

Answer:

a = 5, and b = - 51