Solve -15x^2-35x+100

$$- 15 x ^ { 2 } - 35 x + 100$$

Factor

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

Solution Steps

Factor out 5.

$$5\left(-3x^{2}-7x+20\right)$$

Consider -3x^{2}-7x+20. Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-3\times 20=-60$$

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 -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -7.

$$a=5 b=-12$$

Rewrite -3x^{2}-7x+20 as \left(-3x^{2}+5x\right)+\left(-12x+20\right).

$$\left(-3x^{2}+5x\right)+\left(-12x+20\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$-15x^{2}-35x+100=0$$

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.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\left(-15\right)\times 100}}{2\left(-15\right)}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\left(-15\right)\times 100}}{2\left(-15\right)}$$

Multiply -4 times -15.

$$x=\frac{-\left(-35\right)±\sqrt{1225+60\times 100}}{2\left(-15\right)}$$

Multiply 60 times 100.

$$x=\frac{-\left(-35\right)±\sqrt{1225+6000}}{2\left(-15\right)}$$

Add 1225 to 6000.

$$x=\frac{-\left(-35\right)±\sqrt{7225}}{2\left(-15\right)}$$

Take the square root of 7225.

$$x=\frac{-\left(-35\right)±85}{2\left(-15\right)}$$

The opposite of -35 is 35.

$$x=\frac{35±85}{2\left(-15\right)}$$

Multiply 2 times -15.

$$x=\frac{35±85}{-30}$$

Now solve the equation x=\frac{35±85}{-30} when ± is plus. Add 35 to 85.

$$x=\frac{120}{-30}$$

Divide 120 by -30.

$$x=-4$$

Now solve the equation x=\frac{35±85}{-30} when ± is minus. Subtract 85 from 35.

$$x=-\frac{50}{-30}$$

Reduce the fraction \frac{-50}{-30} to lowest terms by extracting and canceling out 10.

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

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{5}{3} for x_{2}.

$$-15x^{2}-35x+100=-15\left(x-\left(-4\right)\right)\left(x-\frac{5}{3}\right)$$

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

$$-15x^{2}-35x+100=-15\left(x+4\right)\left(x-\frac{5}{3}\right)$$

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

$$-15x^{2}-35x+100=-15\left(x+4\right)\times \frac{-3x+5}{-3}$$

Cancel out 3, the greatest common factor in -15 and 3.

$$-15x^{2}-35x+100=5\left(x+4\right)\left(-3x+5\right)$$

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.

$$x ^ 2 +\frac{7}{3}x -\frac{20}{3} = 0$$

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

$$r + s = -\frac{7}{3} rs = -\frac{20}{3}$$

Two numbers r and s sum up to -\frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{3} = -\frac{7}{6}. 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>

$$r = -\frac{7}{6} - u s = -\frac{7}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{3}

$$(-\frac{7}{6} - u) (-\frac{7}{6} + u) = -\frac{20}{3}$$

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

$$\frac{49}{36} - u^2 = -\frac{20}{3}$$

Simplify the expression by subtracting \frac{49}{36} on both sides

$$-u^2 = -\frac{20}{3}-\frac{49}{36} = -\frac{289}{36}$$

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

$$u^2 = \frac{289}{36} u = \pm\sqrt{\frac{289}{36}} = \pm \frac{17}{6}$$

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

$$r =-\frac{7}{6} - \frac{17}{6} = -4 s = -\frac{7}{6} + \frac{17}{6} = 1.667$$

Evaluate

$$100-35x-15x^{2}$$