Solve 15b^2-31b+10

$$15 b ^ { 2 } - 31 b + 10$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15b^{2}+pb+qb+10. To find p and q, set up a system to be solved.

$$p+q=-31 pq=15\times 10=150$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 150.

$$-1,-150 -2,-75 -3,-50 -5,-30 -6,-25 -10,-15$$

Calculate the sum for each pair.

$$-1-150=-151 -2-75=-77 -3-50=-53 -5-30=-35 -6-25=-31 -10-15=-25$$

The solution is the pair that gives sum -31.

$$p=-25 q=-6$$

Rewrite 15b^{2}-31b+10 as \left(15b^{2}-25b\right)+\left(-6b+10\right).

$$\left(15b^{2}-25b\right)+\left(-6b+10\right)$$

Factor out 5b in the first and -2 in the second group.

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

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

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

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.

$$15b^{2}-31b+10=0$$

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.

$$b=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 15\times 10}}{2\times 15}$$

Square -31.

$$b=\frac{-\left(-31\right)±\sqrt{961-4\times 15\times 10}}{2\times 15}$$

Multiply -4 times 15.

$$b=\frac{-\left(-31\right)±\sqrt{961-60\times 10}}{2\times 15}$$

Multiply -60 times 10.

$$b=\frac{-\left(-31\right)±\sqrt{961-600}}{2\times 15}$$

Add 961 to -600.

$$b=\frac{-\left(-31\right)±\sqrt{361}}{2\times 15}$$

Take the square root of 361.

$$b=\frac{-\left(-31\right)±19}{2\times 15}$$

The opposite of -31 is 31.

$$b=\frac{31±19}{2\times 15}$$

Multiply 2 times 15.

$$b=\frac{31±19}{30}$$

Now solve the equation b=\frac{31±19}{30} when ± is plus. Add 31 to 19.

$$b=\frac{50}{30}$$

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

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

Now solve the equation b=\frac{31±19}{30} when ± is minus. Subtract 19 from 31.

$$b=\frac{12}{30}$$

Reduce the fraction \frac{12}{30} to lowest terms by extracting and canceling out 6.

$$b=\frac{2}{5}$$

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

$$15b^{2}-31b+10=15\left(b-\frac{5}{3}\right)\left(b-\frac{2}{5}\right)$$

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

$$15b^{2}-31b+10=15\times \frac{3b-5}{3}\left(b-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$15b^{2}-31b+10=15\times \frac{3b-5}{3}\times \frac{5b-2}{5}$$

Multiply \frac{3b-5}{3} times \frac{5b-2}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15b^{2}-31b+10=15\times \frac{\left(3b-5\right)\left(5b-2\right)}{3\times 5}$$

Multiply 3 times 5.

$$15b^{2}-31b+10=15\times \frac{\left(3b-5\right)\left(5b-2\right)}{15}$$

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

$$15b^{2}-31b+10=\left(3b-5\right)\left(5b-2\right)$$

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 15

$$x ^ 2 -\frac{31}{15}x +\frac{2}{3} = 0$$

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

$$r + s = \frac{31}{15} rs = \frac{2}{3}$$

Two numbers r and s sum up to \frac{31}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{31}{15} = \frac{31}{30}. 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>

$$r = \frac{31}{30} - u s = \frac{31}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}

$$(\frac{31}{30} - u) (\frac{31}{30} + u) = \frac{2}{3}$$

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

$$\frac{961}{900} - u^2 = \frac{2}{3}$$

Simplify the expression by subtracting \frac{961}{900} on both sides

$$-u^2 = \frac{2}{3}-\frac{961}{900} = -\frac{361}{900}$$

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

$$u^2 = \frac{361}{900} u = \pm\sqrt{\frac{361}{900}} = \pm \frac{19}{30}$$

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

$$r =\frac{31}{30} - \frac{19}{30} = 0.400 s = \frac{31}{30} + \frac{19}{30} = 1.667$$

Evaluate

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