Find constants a, b, and c given that: 6x^3+7x^2-19x+7=(2x-1)(ax^2+bx+c) for all x.
To find the constants a, b, and c, we need to expand the expression on the right hand side of the equation and compare it to the given polynomial:
(2x - 1)(ax^2 + bx + c) = (2ax^3 + (2bx^2 - ax^2) + (cx - bx) + c) = 2ax^3 + (2b - a)x^2 + (c - b)x + c
Now we can equate the coefficients of the corresponding terms from both sides of the equation:
For x^3 term: 2ax^3 = 6x^3
This gives us: 2a = 6, therefore, a = 3.
For x^2 term: (2b - a)x^2 = 7x^2
Substituting the value of a, we have: (2b - 3)x^2 = 7x^2
This gives us: 2b - 3 = 7, therefore, b = 5.
For x term: (c - b)x = -19x
Substituting the value of b, we have: (c - 5)x = -19x
This gives us: c - 5 = -19, therefore, c = -14.
Therefore, the constants are a = 3, b = 5, and c = -14.
You can just do the long division, or note that
(2x-1)(ax^2+bx+c) = 2ax^3 + (-a+2b)x^2 + (-b+2c)x - c
For this to be identical to 6x^3+7x^2-19x+7, you need
2a = 6
-a+2b = 7
-b+2c = -19
-c = 7
Clearly, a=3, c = -7
and I'm sure you can now find b.