Solve =-3x^2+2x+1text{sebagai}f(x)=a(x-h)^2+k

$$= - 3 x ^ { 2 } + 2 x + 1 \text { sebagai } f ( x ) = a ( x - h ) ^ { 2 } + k$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -3x^{2}+2x+1 as \left(-3x^{2}+3x\right)+\left(-x+1\right).

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

Factor out 3x in -3x^{2}+3x.

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

Factor out common term -x+1 by using distributive property.

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

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.

$$-3x^{2}+2x+1=0$$

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.

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

Square 2.

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

Multiply -4 times -3.

$$x=\frac{-2±\sqrt{4+12}}{2\left(-3\right)}$$

Add 4 to 12.

$$x=\frac{-2±\sqrt{16}}{2\left(-3\right)}$$

Take the square root of 16.

$$x=\frac{-2±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±4}{-6}$$

Now solve the equation x=\frac{-2±4}{-6} when ± is plus. Add -2 to 4.

$$x=\frac{2}{-6}$$

Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{3}$$

Now solve the equation x=\frac{-2±4}{-6} when ± is minus. Subtract 4 from -2.

$$x=-\frac{6}{-6}$$

Divide -6 by -6.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{3} for x_{1} and 1 for x_{2}.

$$-3x^{2}+2x+1=-3\left(x-\left(-\frac{1}{3}\right)\right)\left(x-1\right)$$

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

$$-3x^{2}+2x+1=-3\left(x+\frac{1}{3}\right)\left(x-1\right)$$

Add \frac{1}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}+2x+1=-3\times \frac{-3x-1}{-3}\left(x-1\right)$$

Cancel out 3, the greatest common factor in -3 and 3.

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

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.

$$x ^ 2 -\frac{2}{3}x -\frac{1}{3} = 0$$

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

$$r + s = \frac{2}{3} rs = -\frac{1}{3}$$

Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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>

$$r = \frac{1}{3} - u s = \frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}

$$(\frac{1}{3} - u) (\frac{1}{3} + u) = -\frac{1}{3}$$

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

$$\frac{1}{9} - u^2 = -\frac{1}{3}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{1}{3}-\frac{1}{9} = -\frac{4}{9}$$

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

$$u^2 = \frac{4}{9} u = \pm\sqrt{\frac{4}{9}} = \pm \frac{2}{3}$$

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

$$r =\frac{1}{3} - \frac{2}{3} = -0.333 s = \frac{1}{3} + \frac{2}{3} = 1$$

Evaluate

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