Solve 25b^2-35b-120

$$25 b ^ { 2 } - 35 b - 120$$

Factor

$$5\left(b-3\right)\left(5b+8\right)$$

Solution Steps

Factor out 5.

$$5\left(5b^{2}-7b-24\right)$$

Consider 5b^{2}-7b-24. Factor the expression by grouping. First, the expression needs to be rewritten as 5b^{2}+pb+qb-24. To find p and q, set up a system to be solved.

$$p+q=-7 pq=5\left(-24\right)=-120$$

Since pq is negative, p and q have the opposite signs. Since p+q 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.

$$p=-15 q=8$$

Rewrite 5b^{2}-7b-24 as \left(5b^{2}-15b\right)+\left(8b-24\right).

$$\left(5b^{2}-15b\right)+\left(8b-24\right)$$

Factor out 5b in the first and 8 in the second group.

$$5b\left(b-3\right)+8\left(b-3\right)$$

Factor out common term b-3 by using distributive property.

$$\left(b-3\right)\left(5b+8\right)$$

Rewrite the complete factored expression.

$$5\left(b-3\right)\left(5b+8\right)$$

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.

$$25b^{2}-35b-120=0$$

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.

$$b=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 25\left(-120\right)}}{2\times 25}$$

Square -35.

$$b=\frac{-\left(-35\right)±\sqrt{1225-4\times 25\left(-120\right)}}{2\times 25}$$

Multiply -4 times 25.

$$b=\frac{-\left(-35\right)±\sqrt{1225-100\left(-120\right)}}{2\times 25}$$

Multiply -100 times -120.

$$b=\frac{-\left(-35\right)±\sqrt{1225+12000}}{2\times 25}$$

Add 1225 to 12000.

$$b=\frac{-\left(-35\right)±\sqrt{13225}}{2\times 25}$$

Take the square root of 13225.

$$b=\frac{-\left(-35\right)±115}{2\times 25}$$

The opposite of -35 is 35.

$$b=\frac{35±115}{2\times 25}$$

Multiply 2 times 25.

$$b=\frac{35±115}{50}$$

Now solve the equation b=\frac{35±115}{50} when ± is plus. Add 35 to 115.

$$b=\frac{150}{50}$$

Divide 150 by 50.

$$b=3$$

Now solve the equation b=\frac{35±115}{50} when ± is minus. Subtract 115 from 35.

$$b=-\frac{80}{50}$$

Reduce the fraction \frac{-80}{50} to lowest terms by extracting and canceling out 10.

$$b=-\frac{8}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -\frac{8}{5} for x_{2}.

$$25b^{2}-35b-120=25\left(b-3\right)\left(b-\left(-\frac{8}{5}\right)\right)$$

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

$$25b^{2}-35b-120=25\left(b-3\right)\left(b+\frac{8}{5}\right)$$

Add \frac{8}{5} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25b^{2}-35b-120=25\left(b-3\right)\times \frac{5b+8}{5}$$

Cancel out 5, the greatest common factor in 25 and 5.

$$25b^{2}-35b-120=5\left(b-3\right)\left(5b+8\right)$$

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.This is achieved by dividing both sides of the equation by 25

$$x ^ 2 -\frac{7}{5}x -\frac{24}{5} = 0$$

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

$$r + s = \frac{7}{5} rs = -\frac{24}{5}$$

Two numbers r and s sum up to \frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{5} = \frac{7}{10}. 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>

$$r = \frac{7}{10} - u s = \frac{7}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{24}{5}

$$(\frac{7}{10} - u) (\frac{7}{10} + u) = -\frac{24}{5}$$

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

$$\frac{49}{100} - u^2 = -\frac{24}{5}$$

Simplify the expression by subtracting \frac{49}{100} on both sides

$$-u^2 = -\frac{24}{5}-\frac{49}{100} = -\frac{529}{100}$$

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

$$u^2 = \frac{529}{100} u = \pm\sqrt{\frac{529}{100}} = \pm \frac{23}{10}$$

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

$$r =\frac{7}{10} - \frac{23}{10} = -1.600 s = \frac{7}{10} + \frac{23}{10} = 3$$

Evaluate

$$5\left(b-3\right)\left(5b+8\right)$$