Solve 4x^2-68x+145

$$4 x ^ { 2 } - 68 x + 145$$

Factor

$$\left(2x-29\right)\left(2x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+145. To find a and b, set up a system to be solved.

$$a+b=-68 ab=4\times 145=580$$

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 580.

$$-1,-580 -2,-290 -4,-145 -5,-116 -10,-58 -20,-29$$

Calculate the sum for each pair.

$$-1-580=-581 -2-290=-292 -4-145=-149 -5-116=-121 -10-58=-68 -20-29=-49$$

The solution is the pair that gives sum -68.

$$a=-58 b=-10$$

Rewrite 4x^{2}-68x+145 as \left(4x^{2}-58x\right)+\left(-10x+145\right).

$$\left(4x^{2}-58x\right)+\left(-10x+145\right)$$

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

$$2x\left(2x-29\right)-5\left(2x-29\right)$$

Factor out common term 2x-29 by using distributive property.

$$\left(2x-29\right)\left(2x-5\right)$$

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.

$$4x^{2}-68x+145=0$$

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.

$$x=\frac{-\left(-68\right)±\sqrt{\left(-68\right)^{2}-4\times 4\times 145}}{2\times 4}$$

Square -68.

$$x=\frac{-\left(-68\right)±\sqrt{4624-4\times 4\times 145}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-68\right)±\sqrt{4624-16\times 145}}{2\times 4}$$

Multiply -16 times 145.

$$x=\frac{-\left(-68\right)±\sqrt{4624-2320}}{2\times 4}$$

Add 4624 to -2320.

$$x=\frac{-\left(-68\right)±\sqrt{2304}}{2\times 4}$$

Take the square root of 2304.

$$x=\frac{-\left(-68\right)±48}{2\times 4}$$

The opposite of -68 is 68.

$$x=\frac{68±48}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{68±48}{8}$$

Now solve the equation x=\frac{68±48}{8} when ± is plus. Add 68 to 48.

$$x=\frac{116}{8}$$

Reduce the fraction \frac{116}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{29}{2}$$

Now solve the equation x=\frac{68±48}{8} when ± is minus. Subtract 48 from 68.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{29}{2} for x_{1} and \frac{5}{2} for x_{2}.

$$4x^{2}-68x+145=4\left(x-\frac{29}{2}\right)\left(x-\frac{5}{2}\right)$$

Subtract \frac{29}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-68x+145=4\times \frac{2x-29}{2}\left(x-\frac{5}{2}\right)$$

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

$$4x^{2}-68x+145=4\times \frac{2x-29}{2}\times \frac{2x-5}{2}$$

Multiply \frac{2x-29}{2} times \frac{2x-5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-68x+145=4\times \frac{\left(2x-29\right)\left(2x-5\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}-68x+145=4\times \frac{\left(2x-29\right)\left(2x-5\right)}{4}$$

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

$$4x^{2}-68x+145=\left(2x-29\right)\left(2x-5\right)$$

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 4

$$x ^ 2 -17x +\frac{145}{4} = 0$$

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

$$r + s = 17 rs = \frac{145}{4}$$

Two numbers r and s sum up to 17 exactly when the average of the two numbers is \frac{1}{2}*17 = \frac{17}{2}. 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>

$$r = \frac{17}{2} - u s = \frac{17}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{145}{4}

$$(\frac{17}{2} - u) (\frac{17}{2} + u) = \frac{145}{4}$$

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

$$\frac{289}{4} - u^2 = \frac{145}{4}$$

Simplify the expression by subtracting \frac{289}{4} on both sides

$$-u^2 = \frac{145}{4}-\frac{289}{4} = -36$$

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

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

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

$$r =\frac{17}{2} - 6 = 2.500 s = \frac{17}{2} + 6 = 14.500$$

Evaluate

$$\left(2x-29\right)\left(2x-5\right)$$