Solve 6-7x-5x^2

$$6 - 7 x - 5 x ^ { 2 }$$

Factor

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

Solution Steps

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

$$-5x^{2}-7x+6$$

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

$$a+b=-7 ab=-5\times 6=-30$$

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 -30.

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

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -7.

$$a=3 b=-10$$

Rewrite -5x^{2}-7x+6 as \left(-5x^{2}+3x\right)+\left(-10x+6\right).

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

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

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

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

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

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.

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

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-7\right)±\sqrt{49+20\times 6}}{2\left(-5\right)}$$

Multiply 20 times 6.

$$x=\frac{-\left(-7\right)±\sqrt{49+120}}{2\left(-5\right)}$$

Add 49 to 120.

$$x=\frac{-\left(-7\right)±\sqrt{169}}{2\left(-5\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-7\right)±13}{2\left(-5\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±13}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{7±13}{-10}$$

Now solve the equation x=\frac{7±13}{-10} when ± is plus. Add 7 to 13.

$$x=\frac{20}{-10}$$

Divide 20 by -10.

$$x=-2$$

Now solve the equation x=\frac{7±13}{-10} when ± is minus. Subtract 13 from 7.

$$x=-\frac{6}{-10}$$

Reduce the fraction \frac{-6}{-10} to lowest terms by extracting and canceling out 2.

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

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

$$-5x^{2}-7x+6=-5\left(x-\left(-2\right)\right)\left(x-\frac{3}{5}\right)$$

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

$$-5x^{2}-7x+6=-5\left(x+2\right)\left(x-\frac{3}{5}\right)$$

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

$$-5x^{2}-7x+6=-5\left(x+2\right)\times \frac{-5x+3}{-5}$$

Cancel out 5, the greatest common factor in -5 and 5.

$$-5x^{2}-7x+6=\left(x+2\right)\left(-5x+3\right)$$

Evaluate

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