Solve 10-4x-14x^2

$$10 - 4 x - 14 x ^ { 2 }$$

Factor

$$2\left(-x-1\right)\left(7x-5\right)$$

Solution Steps

Factor out 2.

$$2\left(5-2x-7x^{2}\right)$$

Consider 5-2x-7x^{2}. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

Factor the expression by grouping. First, the expression needs to be rewritten as -7x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-7\times 5=-35$$

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 -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=5 b=-7$$

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

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

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

$$-x\left(7x-5\right)-\left(7x-5\right)$$

Factor out common term 7x-5 by using distributive property.

$$\left(7x-5\right)\left(-x-1\right)$$

Rewrite the complete factored expression.

$$2\left(7x-5\right)\left(-x-1\right)$$

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.

$$-14x^{2}-4x+10=0$$

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.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-14\right)\times 10}}{2\left(-14\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-14\right)\times 10}}{2\left(-14\right)}$$

Multiply -4 times -14.

$$x=\frac{-\left(-4\right)±\sqrt{16+56\times 10}}{2\left(-14\right)}$$

Multiply 56 times 10.

$$x=\frac{-\left(-4\right)±\sqrt{16+560}}{2\left(-14\right)}$$

Add 16 to 560.

$$x=\frac{-\left(-4\right)±\sqrt{576}}{2\left(-14\right)}$$

Take the square root of 576.

$$x=\frac{-\left(-4\right)±24}{2\left(-14\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±24}{2\left(-14\right)}$$

Multiply 2 times -14.

$$x=\frac{4±24}{-28}$$

Now solve the equation x=\frac{4±24}{-28} when ± is plus. Add 4 to 24.

$$x=\frac{28}{-28}$$

Divide 28 by -28.

$$x=-1$$

Now solve the equation x=\frac{4±24}{-28} when ± is minus. Subtract 24 from 4.

$$x=-\frac{20}{-28}$$

Reduce the fraction \frac{-20}{-28} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and \frac{5}{7} for x_{2}.

$$-14x^{2}-4x+10=-14\left(x-\left(-1\right)\right)\left(x-\frac{5}{7}\right)$$

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

$$-14x^{2}-4x+10=-14\left(x+1\right)\left(x-\frac{5}{7}\right)$$

Subtract \frac{5}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-14x^{2}-4x+10=-14\left(x+1\right)\times \frac{-7x+5}{-7}$$

Cancel out 7, the greatest common factor in -14 and 7.

$$-14x^{2}-4x+10=2\left(x+1\right)\left(-7x+5\right)$$

Evaluate

$$10-4x-14x^{2}$$