Solve -10x^2-7x+12

$$- 10 x ^ { 2 } - 7 x + 12$$

Factor

$$\left(4-5x\right)\left(2x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -10x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-10\times 12=-120$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -120.

$$1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12$$

Calculate the sum for each pair.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -7.

$$a=8 b=-15$$

Rewrite -10x^{2}-7x+12 as \left(-10x^{2}+8x\right)+\left(-15x+12\right).

$$\left(-10x^{2}+8x\right)+\left(-15x+12\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(-5x+4\right)+3\left(-5x+4\right)$$

Factor out common term -5x+4 by using distributive property.

$$\left(-5x+4\right)\left(2x+3\right)$$

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.

$$-10x^{2}-7x+12=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-10\right)\times 12}}{2\left(-10\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-10\right)\times 12}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-7\right)±\sqrt{49+40\times 12}}{2\left(-10\right)}$$

Multiply 40 times 12.

$$x=\frac{-\left(-7\right)±\sqrt{49+480}}{2\left(-10\right)}$$

Add 49 to 480.

$$x=\frac{-\left(-7\right)±\sqrt{529}}{2\left(-10\right)}$$

Take the square root of 529.

$$x=\frac{-\left(-7\right)±23}{2\left(-10\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±23}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{7±23}{-20}$$

Now solve the equation x=\frac{7±23}{-20} when ± is plus. Add 7 to 23.

$$x=\frac{30}{-20}$$

Reduce the fraction \frac{30}{-20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{2}$$

Now solve the equation x=\frac{7±23}{-20} when ± is minus. Subtract 23 from 7.

$$x=-\frac{16}{-20}$$

Reduce the fraction \frac{-16}{-20} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{2} for x_{1} and \frac{4}{5} for x_{2}.

$$-10x^{2}-7x+12=-10\left(x-\left(-\frac{3}{2}\right)\right)\left(x-\frac{4}{5}\right)$$

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

$$-10x^{2}-7x+12=-10\left(x+\frac{3}{2}\right)\left(x-\frac{4}{5}\right)$$

Add \frac{3}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-10x^{2}-7x+12=-10\times \frac{-2x-3}{-2}\left(x-\frac{4}{5}\right)$$

Subtract \frac{4}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-10x^{2}-7x+12=-10\times \frac{-2x-3}{-2}\times \frac{-5x+4}{-5}$$

Multiply \frac{-2x-3}{-2} times \frac{-5x+4}{-5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-10x^{2}-7x+12=-10\times \frac{\left(-2x-3\right)\left(-5x+4\right)}{-2\left(-5\right)}$$

Multiply -2 times -5.

$$-10x^{2}-7x+12=-10\times \frac{\left(-2x-3\right)\left(-5x+4\right)}{10}$$

Cancel out 10, the greatest common factor in -10 and 10.

$$-10x^{2}-7x+12=-\left(-2x-3\right)\left(-5x+4\right)$$

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.

$$x ^ 2 +\frac{7}{10}x -\frac{6}{5} = 0$$

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

$$r + s = -\frac{7}{10} rs = -\frac{6}{5}$$

Two numbers r and s sum up to -\frac{7}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{10} = -\frac{7}{20}. 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>

$$r = -\frac{7}{20} - u s = -\frac{7}{20} + u$$

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

$$(-\frac{7}{20} - u) (-\frac{7}{20} + u) = -\frac{6}{5}$$

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

$$\frac{49}{400} - u^2 = -\frac{6}{5}$$

Simplify the expression by subtracting \frac{49}{400} on both sides

$$-u^2 = -\frac{6}{5}-\frac{49}{400} = -\frac{529}{400}$$

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

$$u^2 = \frac{529}{400} u = \pm\sqrt{\frac{529}{400}} = \pm \frac{23}{20}$$

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

$$r =-\frac{7}{20} - \frac{23}{20} = -1.500 s = -\frac{7}{20} + \frac{23}{20} = 0.800$$

Evaluate

$$\left(4-5x\right)\left(2x+3\right)$$