Solve 4x^2-56x-1829

$$4 x ^ { 2 } - 56 x - 1829$$

Factor

$$\left(2x-59\right)\left(2x+31\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-1829. To find a and b, set up a system to be solved.

$$a+b=-56 ab=4\left(-1829\right)=-7316$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -7316.

$$1,-7316 2,-3658 4,-1829 31,-236 59,-124 62,-118$$

Calculate the sum for each pair.

$$1-7316=-7315 2-3658=-3656 4-1829=-1825 31-236=-205 59-124=-65 62-118=-56$$

The solution is the pair that gives sum -56.

$$a=-118 b=62$$

Rewrite 4x^{2}-56x-1829 as \left(4x^{2}-118x\right)+\left(62x-1829\right).

$$\left(4x^{2}-118x\right)+\left(62x-1829\right)$$

Factor out 2x in the first and 31 in the second group.

$$2x\left(2x-59\right)+31\left(2x-59\right)$$

Factor out common term 2x-59 by using distributive property.

$$\left(2x-59\right)\left(2x+31\right)$$

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.

$$4x^{2}-56x-1829=0$$

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.

$$x=\frac{-\left(-56\right)±\sqrt{\left(-56\right)^{2}-4\times 4\left(-1829\right)}}{2\times 4}$$

Square -56.

$$x=\frac{-\left(-56\right)±\sqrt{3136-4\times 4\left(-1829\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-56\right)±\sqrt{3136-16\left(-1829\right)}}{2\times 4}$$

Multiply -16 times -1829.

$$x=\frac{-\left(-56\right)±\sqrt{3136+29264}}{2\times 4}$$

Add 3136 to 29264.

$$x=\frac{-\left(-56\right)±\sqrt{32400}}{2\times 4}$$

Take the square root of 32400.

$$x=\frac{-\left(-56\right)±180}{2\times 4}$$

The opposite of -56 is 56.

$$x=\frac{56±180}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{56±180}{8}$$

Now solve the equation x=\frac{56±180}{8} when ± is plus. Add 56 to 180.

$$x=\frac{236}{8}$$

Reduce the fraction \frac{236}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{59}{2}$$

Now solve the equation x=\frac{56±180}{8} when ± is minus. Subtract 180 from 56.

$$x=-\frac{124}{8}$$

Reduce the fraction \frac{-124}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{31}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{59}{2} for x_{1} and -\frac{31}{2} for x_{2}.

$$4x^{2}-56x-1829=4\left(x-\frac{59}{2}\right)\left(x-\left(-\frac{31}{2}\right)\right)$$

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

$$4x^{2}-56x-1829=4\left(x-\frac{59}{2}\right)\left(x+\frac{31}{2}\right)$$

Subtract \frac{59}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-56x-1829=4\times \frac{2x-59}{2}\left(x+\frac{31}{2}\right)$$

Add \frac{31}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-56x-1829=4\times \frac{2x-59}{2}\times \frac{2x+31}{2}$$

Multiply \frac{2x-59}{2} times \frac{2x+31}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-56x-1829=4\times \frac{\left(2x-59\right)\left(2x+31\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}-56x-1829=4\times \frac{\left(2x-59\right)\left(2x+31\right)}{4}$$

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

$$4x^{2}-56x-1829=\left(2x-59\right)\left(2x+31\right)$$

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

$$x ^ 2 -14x -\frac{1829}{4} = 0$$

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

$$r + s = 14 rs = -\frac{1829}{4}$$

Two numbers r and s sum up to 14 exactly when the average of the two numbers is \frac{1}{2}*14 = 7. 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>

$$r = 7 - u s = 7 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1829}{4}

$$(7 - u) (7 + u) = -\frac{1829}{4}$$

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

$$49 - u^2 = -\frac{1829}{4}$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = -\frac{1829}{4}-49 = -\frac{2025}{4}$$

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

$$u^2 = \frac{2025}{4} u = \pm\sqrt{\frac{2025}{4}} = \pm \frac{45}{2}$$

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

$$r =7 - \frac{45}{2} = -15.500 s = 7 + \frac{45}{2} = 29.500$$

Evaluate

$$\left(2x-59\right)\left(2x+31\right)$$