Dividend: 4x^4 -x^2 +17x^2 +11x +4 Divisor: 4x +3 Quotient: x^3 -x^2 +5x -1 Remainder: 7
To find the extreme value of the parabola y = x2 - 4x + 3 ...(1) Take the derivative of the equation.y = x2 - 4x + 3y' = 2x - 4(2) Set the derivative = 0 and solve for x.y' = 2x - 40 = 2x - 42x = 4x = 4/2x = 2(3) Plug this x value back into the original equation to find the associated y coordinate.x = 2y = x2 - 4x + 3y = (2)2 - 4(2) + 3y = 4 - 8 + 3y = -1So the vertex is at (2, -1).
Dividend: x3+4x2-9x-36 Divisor: x+3 Quotient: x2+x-12
-17
A graph that has 1 parabolla that has a minimum and 1 positive line.
4x + 3 doesn't factor. It is in its simplest form.
x2-6x+9 = (x-3)(x-3) when factorised.
It is 2(2x+3) when factorised
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
You can work this out with long division, by checking to see if (x2 - 1) is a factor of (2x4 + 4x3 - x2 + 4x - 3). It is. Unfortunately, the WikiAnswers system is somewhat limited in depicting things such as long division, so we won't be able to represent it here. In short though, (2x4 + 4x3 - x2 + 4x - 3) / (x2 + 1) is equal to 2x2 + 4x - 3. which means that: (x2 + 1) / (2x4 + 4x3 - x2 + 4x - 3) = (x2 + 1) / (x2 + 1)(2x2 + 4x - 3) = 1 / (2x2 + 4x - 3)
y = 4x-3
x2-9x+18 = (x-3)(x-6) when factorised
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2 - 4x + 3 is already in standard form.