Solve 2u^2-11u+15

$$2 u ^ { 2 } - 11 u + 15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2u^{2}+au+bu+15. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 15=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -11.

$$a=-6 b=-5$$

Rewrite 2u^{2}-11u+15 as \left(2u^{2}-6u\right)+\left(-5u+15\right).

$$\left(2u^{2}-6u\right)+\left(-5u+15\right)$$

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

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

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

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

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.

$$2u^{2}-11u+15=0$$

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.

$$u=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 15}}{2\times 2}$$

Square -11.

$$u=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 15}}{2\times 2}$$

Multiply -4 times 2.

$$u=\frac{-\left(-11\right)±\sqrt{121-8\times 15}}{2\times 2}$$

Multiply -8 times 15.

$$u=\frac{-\left(-11\right)±\sqrt{121-120}}{2\times 2}$$

Add 121 to -120.

$$u=\frac{-\left(-11\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$u=\frac{-\left(-11\right)±1}{2\times 2}$$

The opposite of -11 is 11.

$$u=\frac{11±1}{2\times 2}$$

Multiply 2 times 2.

$$u=\frac{11±1}{4}$$

Now solve the equation u=\frac{11±1}{4} when ± is plus. Add 11 to 1.

$$u=\frac{12}{4}$$

Divide 12 by 4.

$$u=3$$

Now solve the equation u=\frac{11±1}{4} when ± is minus. Subtract 1 from 11.

$$u=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

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

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

$$2u^{2}-11u+15=2\left(u-3\right)\left(u-\frac{5}{2}\right)$$

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

$$2u^{2}-11u+15=2\left(u-3\right)\times \frac{2u-5}{2}$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2u^{2}-11u+15=\left(u-3\right)\left(2u-5\right)$$

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.This is achieved by dividing both sides of the equation by 2

$$x ^ 2 -\frac{11}{2}x +\frac{15}{2} = 0$$

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

$$r + s = \frac{11}{2} rs = \frac{15}{2}$$

Two numbers r and s sum up to \frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{2} = \frac{11}{4}. 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>

$$r = \frac{11}{4} - u s = \frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{2}

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = \frac{15}{2}$$

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

$$\frac{121}{16} - u^2 = \frac{15}{2}$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = \frac{15}{2}-\frac{121}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =\frac{11}{4} - \frac{1}{4} = 2.500 s = \frac{11}{4} + \frac{1}{4} = 3$$

Evaluate

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