Polynomial x²-6y²-xy-5x-5y+6 was factored into (x+ay+b)(x+cy-2) for constants a, b and c. Find the value of a+bc

We know that the polynomial:

x²- 6y²- xy - 5x - 5y + 6

is rewritten as:

(x+ay+b)(x+cy-2)

First, lets expand the above expression:

x^2 + x*(cy) + x*(-2) + (ay)*x + (ay)*(cy) + (ay)*(-2) + b*x + b*(cy) - 2*b

Now we can simplify this to get:

x² + c*(xy) - 2*x + a*(xy) + ac*y² - 2a*y + b*x + bc*y  - 2b

now let's group together the terms with the same variables:

x² + (c + a)*(xy) + (b - 2)*x + (bc - 2a)*y + ac*y² - 2b

And that must be equal to:

x²- 6y²- xy - 5x - 5y + 6

notice that equations are equal if and only if all the correspondent factors are equal.

notice that in both cases, the factor that multiplies the x² term is 1.

for the y² term we will have:

a*c = -6

for the xy term we will have

c + a = -1

for the x term we will have

b - 2  = - 5

for the y term we will have

bc - 2a = -5

for the constant term, we will have:

-2b = 6

Then we have a lot of equations, rewriting these we have:

a*c = -6

c + a = -1

b - 2  = -5

bc - 2a = -5

-2b = 6

From the fourth equation, b - 2  = -5

we can get:

b = -5 + 2 = -3

b = -3

notice that for the last equation:

-2b = 6

b = 6/-2 = -3

we have the same solution

Then we can replace the value of b in the above equations to get:

a*c = -6

c + a = -1

-3*c - 2a = -5

Now, we need to isolate one of the variables in one of the equations.

For example, we can isolate c in the second one to get:

c = -1 - a

now we can replace that in other equation, for example the third one:

-3*(-1 - a) - 2a = -5

now we can solve that for a.

3 + 3a - 2a = -5

3 + (3 - 2)a = -5

3 + a = -5

a =  -5 - 3 = -8

a = -8

now we can use the equation "c = -1 - a" to find the value of c:

c = -1 -(-8) = -1 + 8 = 7

c = 7

then we have:

b = -3

a = -8

c = 7

then:

a + b*c = -8 + (-3)*7 = -8 - 21 = -29

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!