Solve 25p^2-74p-3

$$25 p ^ { 2 } - 74 p - 3$$

Factor

$$\left(p-3\right)\left(25p+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25p^{2}+ap+bp-3. To find a and b, set up a system to be solved.

$$a+b=-74 ab=25\left(-3\right)=-75$$

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 -75.

$$1,-75 3,-25 5,-15$$

Calculate the sum for each pair.

$$1-75=-74 3-25=-22 5-15=-10$$

The solution is the pair that gives sum -74.

$$a=-75 b=1$$

Rewrite 25p^{2}-74p-3 as \left(25p^{2}-75p\right)+\left(p-3\right).

$$\left(25p^{2}-75p\right)+\left(p-3\right)$$

Factor out 25p in 25p^{2}-75p.

$$25p\left(p-3\right)+p-3$$

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

$$\left(p-3\right)\left(25p+1\right)$$

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.

$$25p^{2}-74p-3=0$$

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.

$$p=\frac{-\left(-74\right)±\sqrt{\left(-74\right)^{2}-4\times 25\left(-3\right)}}{2\times 25}$$

Square -74.

$$p=\frac{-\left(-74\right)±\sqrt{5476-4\times 25\left(-3\right)}}{2\times 25}$$

Multiply -4 times 25.

$$p=\frac{-\left(-74\right)±\sqrt{5476-100\left(-3\right)}}{2\times 25}$$

Multiply -100 times -3.

$$p=\frac{-\left(-74\right)±\sqrt{5476+300}}{2\times 25}$$

Add 5476 to 300.

$$p=\frac{-\left(-74\right)±\sqrt{5776}}{2\times 25}$$

Take the square root of 5776.

$$p=\frac{-\left(-74\right)±76}{2\times 25}$$

The opposite of -74 is 74.

$$p=\frac{74±76}{2\times 25}$$

Multiply 2 times 25.

$$p=\frac{74±76}{50}$$

Now solve the equation p=\frac{74±76}{50} when ± is plus. Add 74 to 76.

$$p=\frac{150}{50}$$

Divide 150 by 50.

$$p=3$$

Now solve the equation p=\frac{74±76}{50} when ± is minus. Subtract 76 from 74.

$$p=-\frac{2}{50}$$

Reduce the fraction \frac{-2}{50} to lowest terms by extracting and canceling out 2.

$$p=-\frac{1}{25}$$

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{1}{25} for x_{2}.

$$25p^{2}-74p-3=25\left(p-3\right)\left(p-\left(-\frac{1}{25}\right)\right)$$

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

$$25p^{2}-74p-3=25\left(p-3\right)\left(p+\frac{1}{25}\right)$$

Add \frac{1}{25} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25p^{2}-74p-3=25\left(p-3\right)\times \frac{25p+1}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25p^{2}-74p-3=\left(p-3\right)\left(25p+1\right)$$

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 25

$$x ^ 2 -\frac{74}{25}x -\frac{3}{25} = 0$$

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

$$r + s = \frac{74}{25} rs = -\frac{3}{25}$$

Two numbers r and s sum up to \frac{74}{25} exactly when the average of the two numbers is \frac{1}{2}*\frac{74}{25} = \frac{37}{25}. 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>

$$r = \frac{37}{25} - u s = \frac{37}{25} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{25}

$$(\frac{37}{25} - u) (\frac{37}{25} + u) = -\frac{3}{25}$$

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

$$\frac{1369}{625} - u^2 = -\frac{3}{25}$$

Simplify the expression by subtracting \frac{1369}{625} on both sides

$$-u^2 = -\frac{3}{25}-\frac{1369}{625} = -\frac{1444}{625}$$

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

$$u^2 = \frac{1444}{625} u = \pm\sqrt{\frac{1444}{625}} = \pm \frac{38}{25}$$

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

$$r =\frac{37}{25} - \frac{38}{25} = -0.040 s = \frac{37}{25} + \frac{38}{25} = 3$$

Evaluate

$$\left(p-3\right)\left(25p+1\right)$$