Solve -3y^2-6y-3

$$- 3 y ^ { 2 } - 6 y - 3$$

Factor

$$-3\left(y+1\right)^{2}$$

Solution Steps

Factor out 3.

$$3\left(-y^{2}-2y-1\right)$$

Consider -y^{2}-2y-1. Factor the expression by grouping. First, the expression needs to be rewritten as -y^{2}+ay+by-1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-1$$

Rewrite -y^{2}-2y-1 as \left(-y^{2}-y\right)+\left(-y-1\right).

$$\left(-y^{2}-y\right)+\left(-y-1\right)$$

Factor out -y in the first and -1 in the second group.

$$-y\left(y+1\right)-\left(y+1\right)$$

Factor out common term y+1 by using distributive property.

$$\left(y+1\right)\left(-y-1\right)$$

Rewrite the complete factored expression.

$$3\left(y+1\right)\left(-y-1\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3y^{2}-6y-3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Square -6.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-\left(-6\right)±\sqrt{36+12\left(-3\right)}}{2\left(-3\right)}$$

Multiply 12 times -3.

$$y=\frac{-\left(-6\right)±\sqrt{36-36}}{2\left(-3\right)}$$

Add 36 to -36.

$$y=\frac{-\left(-6\right)±\sqrt{0}}{2\left(-3\right)}$$

Take the square root of 0.

$$y=\frac{-\left(-6\right)±0}{2\left(-3\right)}$$

The opposite of -6 is 6.

$$y=\frac{6±0}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{6±0}{-6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -1 for x_{2}.

$$-3y^{2}-6y-3=-3\left(y-\left(-1\right)\right)\left(y-\left(-1\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3y^{2}-6y-3=-3\left(y+1\right)\left(y+1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +2x +1 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -2 rs = 1$$

Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(-1 - u) (-1 + u) = 1$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$1 - u^2 = 1$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = 1-1 = 0$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 0 u = 0$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r = s = -1$$

Evaluate

$$-3\left(y+1\right)^{2}$$