Solve 4w^2-23w+15=

$$4 w ^ { 2 } - 23 w + 15 =$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4w^{2}+aw+bw+15. To find a and b, set up a system to be solved.

$$a+b=-23 ab=4\times 15=60$$

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 60.

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

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -23.

$$a=-20 b=-3$$

Rewrite 4w^{2}-23w+15 as \left(4w^{2}-20w\right)+\left(-3w+15\right).

$$\left(4w^{2}-20w\right)+\left(-3w+15\right)$$

Factor out 4w in the first and -3 in the second group.

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

Factor out common term w-5 by using distributive property.

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

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.

$$4w^{2}-23w+15=0$$

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.

$$w=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 4\times 15}}{2\times 4}$$

Square -23.

$$w=\frac{-\left(-23\right)±\sqrt{529-4\times 4\times 15}}{2\times 4}$$

Multiply -4 times 4.

$$w=\frac{-\left(-23\right)±\sqrt{529-16\times 15}}{2\times 4}$$

Multiply -16 times 15.

$$w=\frac{-\left(-23\right)±\sqrt{529-240}}{2\times 4}$$

Add 529 to -240.

$$w=\frac{-\left(-23\right)±\sqrt{289}}{2\times 4}$$

Take the square root of 289.

$$w=\frac{-\left(-23\right)±17}{2\times 4}$$

The opposite of -23 is 23.

$$w=\frac{23±17}{2\times 4}$$

Multiply 2 times 4.

$$w=\frac{23±17}{8}$$

Now solve the equation w=\frac{23±17}{8} when ± is plus. Add 23 to 17.

$$w=\frac{40}{8}$$

Divide 40 by 8.

$$w=5$$

Now solve the equation w=\frac{23±17}{8} when ± is minus. Subtract 17 from 23.

$$w=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$w=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and \frac{3}{4} for x_{2}.

$$4w^{2}-23w+15=4\left(w-5\right)\left(w-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from w by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4w^{2}-23w+15=4\left(w-5\right)\times \frac{4w-3}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4w^{2}-23w+15=\left(w-5\right)\left(4w-3\right)$$

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 4

$$x ^ 2 -\frac{23}{4}x +\frac{15}{4} = 0$$

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

$$r + s = \frac{23}{4} rs = \frac{15}{4}$$

Two numbers r and s sum up to \frac{23}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{4} = \frac{23}{8}. 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>

$$r = \frac{23}{8} - u s = \frac{23}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{15}{4}

$$(\frac{23}{8} - u) (\frac{23}{8} + u) = \frac{15}{4}$$

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

$$\frac{529}{64} - u^2 = \frac{15}{4}$$

Simplify the expression by subtracting \frac{529}{64} on both sides

$$-u^2 = \frac{15}{4}-\frac{529}{64} = -\frac{289}{64}$$

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

$$u^2 = \frac{289}{64} u = \pm\sqrt{\frac{289}{64}} = \pm \frac{17}{8}$$

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

$$r =\frac{23}{8} - \frac{17}{8} = 0.750 s = \frac{23}{8} + \frac{17}{8} = 5$$

Evaluate

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