Solve f(x)=-20*x^2-106*x-104

$$f ( x ) = - 20 \cdot x ^ { 2 } - 106 \cdot x - 104$$

Factor

$$2\left(-x-4\right)\left(10x+13\right)$$

Solution Steps

Factor out 2.

$$2\left(-10x^{2}-53x-52\right)$$

Consider -10x^{2}-53x-52. Factor the expression by grouping. First, the expression needs to be rewritten as -10x^{2}+ax+bx-52. To find a and b, set up a system to be solved.

$$a+b=-53 ab=-10\left(-52\right)=520$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 520.

$$-1,-520 -2,-260 -4,-130 -5,-104 -8,-65 -10,-52 -13,-40 -20,-26$$

Calculate the sum for each pair.

$$-1-520=-521 -2-260=-262 -4-130=-134 -5-104=-109 -8-65=-73 -10-52=-62 -13-40=-53 -20-26=-46$$

The solution is the pair that gives sum -53.

$$a=-13 b=-40$$

Rewrite -10x^{2}-53x-52 as \left(-10x^{2}-13x\right)+\left(-40x-52\right).

$$\left(-10x^{2}-13x\right)+\left(-40x-52\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(10x+13\right)-4\left(10x+13\right)$$

Factor out common term 10x+13 by using distributive property.

$$\left(10x+13\right)\left(-x-4\right)$$

Rewrite the complete factored expression.

$$2\left(10x+13\right)\left(-x-4\right)$$

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.

$$-20x^{2}-106x-104=0$$

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.

$$x=\frac{-\left(-106\right)±\sqrt{\left(-106\right)^{2}-4\left(-20\right)\left(-104\right)}}{2\left(-20\right)}$$

Square -106.

$$x=\frac{-\left(-106\right)±\sqrt{11236-4\left(-20\right)\left(-104\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-\left(-106\right)±\sqrt{11236+80\left(-104\right)}}{2\left(-20\right)}$$

Multiply 80 times -104.

$$x=\frac{-\left(-106\right)±\sqrt{11236-8320}}{2\left(-20\right)}$$

Add 11236 to -8320.

$$x=\frac{-\left(-106\right)±\sqrt{2916}}{2\left(-20\right)}$$

Take the square root of 2916.

$$x=\frac{-\left(-106\right)±54}{2\left(-20\right)}$$

The opposite of -106 is 106.

$$x=\frac{106±54}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{106±54}{-40}$$

Now solve the equation x=\frac{106±54}{-40} when ± is plus. Add 106 to 54.

$$x=\frac{160}{-40}$$

Divide 160 by -40.

$$x=-4$$

Now solve the equation x=\frac{106±54}{-40} when ± is minus. Subtract 54 from 106.

$$x=\frac{52}{-40}$$

Reduce the fraction \frac{52}{-40} to lowest terms by extracting and canceling out 4.

$$x=-\frac{13}{10}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and -\frac{13}{10} for x_{2}.

$$-20x^{2}-106x-104=-20\left(x-\left(-4\right)\right)\left(x-\left(-\frac{13}{10}\right)\right)$$

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

$$-20x^{2}-106x-104=-20\left(x+4\right)\left(x+\frac{13}{10}\right)$$

Add \frac{13}{10} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}-106x-104=-20\left(x+4\right)\times \frac{-10x-13}{-10}$$

Cancel out 10, the greatest common factor in -20 and 10.

$$-20x^{2}-106x-104=2\left(x+4\right)\left(-10x-13\right)$$

Evaluate

$$-20x^{2}-106x-104$$