# Given that (x-2) and (x+3) are factors of 2x^4-ax^3-10x^2+bx-54, find a and b.

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Since x-2 and x+3 are factors of f(x) = 2x^4-ax^3-10x^2+bx-54, we can write

f(x) =(x-2)(x+3)Q(x).Or

2x^4-ax^3-10x^2+bx-54 = (x-2)(x+3)Q(x)......(1).

Putting x=2 in (1) we get: 2*2^4-a(2)^3-10(2)^2+b(2)-54 = (2-2)(2+2)Q(2) =0. Or

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

-8a+2b=62 Or

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

Put x=-3. Then

2(-3)^4-a(-3)^3-10(-3)^2+b(-3)-54 = (-2-2)(-2+2)Q(-2) =0. Or

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

27a-3b= -18. Or dividing by 3,

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

Adding eq(2) and eq(3) we get: -4a+9a = 31-6 =25. Or

5a=25. So, **a = 5.**

Substituting in eq (3), we get: 9*5-b = -6. Or

**b **= 9*5+6 = **51.**

We'll apply the division of polynomials. Considering the fact that x-2 and x+3 are factors, that means that 2 and -3 are the roots of the polynomial 2x^4-ax^3-10x^2+bx-54. We know that we could write a polynomial as a product of linear factors, depending on it's roots.

2x^4-ax^3-10x^2+bx-54=(x-2)(x+3)(cx^2+dx+e)

We've noticed that multiplying (x-2)(x+3), we'll obtain a second degree polynomial. But the given polynomial has the fourth degree, so we have to multiply (x-2)(x+3) with another polynomial of second degree.

We'll do the math, to the right side and the result will be:

(x-2)(x+3) = x^2+x-6

(x-2)(x+3)(cx^2+dx+e) = (x^2+x-6)(cx^2+dx+e)

(x^2+x-6)(cx^2+dx+e) = cx^4+dx^3+ex^2+cx^3+dx^2+ex-6cx^2-6dx-6e

If 2 polynomials are identically, that means that the corresponding coefficients are equal.

2x^4= cx^4, so **c=2**

-ax^3=(d+c)x^3, so d+c=-a, where c=2, d+2=-a

-10x^2=(e+d-6c)x^2

-10=e+d-6c

e+d=-10+12

e+d=2, where e=8, so d=2-8, **d=-6**

bx=(e-6d)x

b=e-6d

-6e=-54

**e=8**

d+2=-a, a=-2-d, where d=-6

a=-2-(-6)

a=6-2

**a=4**

b=e-6d, where d=-6 and e=8

b=8-6(-6)

b=8+36

**b=44**