Solve 12-7x-10x^2

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

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

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

Evaluate

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