Solve -6x^2+13x-5?

$$- 6 x ^ { 2 } + 13 x - 5 ?$$

Factor

$$\left(5-3x\right)\left(2x-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -6x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=13 ab=-6\left(-5\right)=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=10 b=3$$

Rewrite -6x^{2}+13x-5 as \left(-6x^{2}+10x\right)+\left(3x-5\right).

$$\left(-6x^{2}+10x\right)+\left(3x-5\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(-3x+5\right)-\left(-3x+5\right)$$

Factor out common term -3x+5 by using distributive property.

$$\left(-3x+5\right)\left(2x-1\right)$$

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.

$$-6x^{2}+13x-5=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-6\right)\left(-5\right)}}{2\left(-6\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-6\right)\left(-5\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-13±\sqrt{169+24\left(-5\right)}}{2\left(-6\right)}$$

Multiply 24 times -5.

$$x=\frac{-13±\sqrt{169-120}}{2\left(-6\right)}$$

Add 169 to -120.

$$x=\frac{-13±\sqrt{49}}{2\left(-6\right)}$$

Take the square root of 49.

$$x=\frac{-13±7}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-13±7}{-12}$$

Now solve the equation x=\frac{-13±7}{-12} when ± is plus. Add -13 to 7.

$$x=-\frac{6}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-13±7}{-12} when ± is minus. Subtract 7 from -13.

$$x=-\frac{20}{-12}$$

Reduce the fraction \frac{-20}{-12} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and \frac{5}{3} for x_{2}.

$$-6x^{2}+13x-5=-6\left(x-\frac{1}{2}\right)\left(x-\frac{5}{3}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}+13x-5=-6\times \frac{-2x+1}{-2}\left(x-\frac{5}{3}\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}+13x-5=-6\times \frac{-2x+1}{-2}\times \frac{-3x+5}{-3}$$

Multiply \frac{-2x+1}{-2} times \frac{-3x+5}{-3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}+13x-5=-6\times \frac{\left(-2x+1\right)\left(-3x+5\right)}{-2\left(-3\right)}$$

Multiply -2 times -3.

$$-6x^{2}+13x-5=-6\times \frac{\left(-2x+1\right)\left(-3x+5\right)}{6}$$

Cancel out 6, the greatest common factor in -6 and 6.

$$-6x^{2}+13x-5=-\left(-2x+1\right)\left(-3x+5\right)$$

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.

$$x ^ 2 -\frac{13}{6}x +\frac{5}{6} = 0$$

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

$$r + s = \frac{13}{6} rs = \frac{5}{6}$$

Two numbers r and s sum up to \frac{13}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{6} = \frac{13}{12}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{13}{12} - u s = \frac{13}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{6}

$$(\frac{13}{12} - u) (\frac{13}{12} + u) = \frac{5}{6}$$

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

$$\frac{169}{144} - u^2 = \frac{5}{6}$$

Simplify the expression by subtracting \frac{169}{144} on both sides

$$-u^2 = \frac{5}{6}-\frac{169}{144} = -\frac{49}{144}$$

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

$$u^2 = \frac{49}{144} u = \pm\sqrt{\frac{49}{144}} = \pm \frac{7}{12}$$

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

$$r =\frac{13}{12} - \frac{7}{12} = 0.500 s = \frac{13}{12} + \frac{7}{12} = 1.667$$

Evaluate

$$\left(5-3x\right)\left(2x-1\right)$$