Solve 4p^2+164p+81

$$4 p ^ { 2 } + 164 p + 81$$

Factor

$$\left(2p+1\right)\left(2p+81\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4p^{2}+ap+bp+81. To find a and b, set up a system to be solved.

$$a+b=164 ab=4\times 81=324$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 324.

$$1,324 2,162 3,108 4,81 6,54 9,36 12,27 18,18$$

Calculate the sum for each pair.

$$1+324=325 2+162=164 3+108=111 4+81=85 6+54=60 9+36=45 12+27=39 18+18=36$$

The solution is the pair that gives sum 164.

$$a=2 b=162$$

Rewrite 4p^{2}+164p+81 as \left(4p^{2}+2p\right)+\left(162p+81\right).

$$\left(4p^{2}+2p\right)+\left(162p+81\right)$$

Factor out 2p in the first and 81 in the second group.

$$2p\left(2p+1\right)+81\left(2p+1\right)$$

Factor out common term 2p+1 by using distributive property.

$$\left(2p+1\right)\left(2p+81\right)$$

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.

$$4p^{2}+164p+81=0$$

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.

$$p=\frac{-164±\sqrt{164^{2}-4\times 4\times 81}}{2\times 4}$$

Square 164.

$$p=\frac{-164±\sqrt{26896-4\times 4\times 81}}{2\times 4}$$

Multiply -4 times 4.

$$p=\frac{-164±\sqrt{26896-16\times 81}}{2\times 4}$$

Multiply -16 times 81.

$$p=\frac{-164±\sqrt{26896-1296}}{2\times 4}$$

Add 26896 to -1296.

$$p=\frac{-164±\sqrt{25600}}{2\times 4}$$

Take the square root of 25600.

$$p=\frac{-164±160}{2\times 4}$$

Multiply 2 times 4.

$$p=\frac{-164±160}{8}$$

Now solve the equation p=\frac{-164±160}{8} when ± is plus. Add -164 to 160.

$$p=-\frac{4}{8}$$

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

$$p=-\frac{1}{2}$$

Now solve the equation p=\frac{-164±160}{8} when ± is minus. Subtract 160 from -164.

$$p=-\frac{324}{8}$$

Reduce the fraction \frac{-324}{8} to lowest terms by extracting and canceling out 4.

$$p=-\frac{81}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{2} for x_{1} and -\frac{81}{2} for x_{2}.

$$4p^{2}+164p+81=4\left(p-\left(-\frac{1}{2}\right)\right)\left(p-\left(-\frac{81}{2}\right)\right)$$

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

$$4p^{2}+164p+81=4\left(p+\frac{1}{2}\right)\left(p+\frac{81}{2}\right)$$

Add \frac{1}{2} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4p^{2}+164p+81=4\times \frac{2p+1}{2}\left(p+\frac{81}{2}\right)$$

Add \frac{81}{2} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4p^{2}+164p+81=4\times \frac{2p+1}{2}\times \frac{2p+81}{2}$$

Multiply \frac{2p+1}{2} times \frac{2p+81}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4p^{2}+164p+81=4\times \frac{\left(2p+1\right)\left(2p+81\right)}{2\times 2}$$

Multiply 2 times 2.

$$4p^{2}+164p+81=4\times \frac{\left(2p+1\right)\left(2p+81\right)}{4}$$

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

$$4p^{2}+164p+81=\left(2p+1\right)\left(2p+81\right)$$

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

$$x ^ 2 +41x +\frac{81}{4} = 0$$

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

$$r + s = -41 rs = \frac{81}{4}$$

Two numbers r and s sum up to -41 exactly when the average of the two numbers is \frac{1}{2}*-41 = -\frac{41}{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>

$$r = -\frac{41}{2} - u s = -\frac{41}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{81}{4}

$$(-\frac{41}{2} - u) (-\frac{41}{2} + u) = \frac{81}{4}$$

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

$$\frac{1681}{4} - u^2 = \frac{81}{4}$$

Simplify the expression by subtracting \frac{1681}{4} on both sides

$$-u^2 = \frac{81}{4}-\frac{1681}{4} = -400$$

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

$$u^2 = 400 u = \pm\sqrt{400} = \pm 20$$

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

$$r =-\frac{41}{2} - 20 = -40.500 s = -\frac{41}{2} + 20 = -0.500$$

Evaluate

$$\left(2p+1\right)\left(2p+81\right)$$