Answer:
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
