Solve 21p^2+120p+75

$$21 p ^ { 2 } + 120 p + 75$$

Factor

$$3\left(p+5\right)\left(7p+5\right)$$

Solution Steps

Factor out 3.

$$3\left(7p^{2}+40p+25\right)$$

Consider 7p^{2}+40p+25. Factor the expression by grouping. First, the expression needs to be rewritten as 7p^{2}+ap+bp+25. To find a and b, set up a system to be solved.

$$a+b=40 ab=7\times 25=175$$

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 175.

$$1,175 5,35 7,25$$

Calculate the sum for each pair.

$$1+175=176 5+35=40 7+25=32$$

The solution is the pair that gives sum 40.

$$a=5 b=35$$

Rewrite 7p^{2}+40p+25 as \left(7p^{2}+5p\right)+\left(35p+25\right).

$$\left(7p^{2}+5p\right)+\left(35p+25\right)$$

Factor out p in the first and 5 in the second group.

$$p\left(7p+5\right)+5\left(7p+5\right)$$

Factor out common term 7p+5 by using distributive property.

$$\left(7p+5\right)\left(p+5\right)$$

Rewrite the complete factored expression.

$$3\left(7p+5\right)\left(p+5\right)$$

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.

$$21p^{2}+120p+75=0$$

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.

$$p=\frac{-120±\sqrt{120^{2}-4\times 21\times 75}}{2\times 21}$$

Square 120.

$$p=\frac{-120±\sqrt{14400-4\times 21\times 75}}{2\times 21}$$

Multiply -4 times 21.

$$p=\frac{-120±\sqrt{14400-84\times 75}}{2\times 21}$$

Multiply -84 times 75.

$$p=\frac{-120±\sqrt{14400-6300}}{2\times 21}$$

Add 14400 to -6300.

$$p=\frac{-120±\sqrt{8100}}{2\times 21}$$

Take the square root of 8100.

$$p=\frac{-120±90}{2\times 21}$$

Multiply 2 times 21.

$$p=\frac{-120±90}{42}$$

Now solve the equation p=\frac{-120±90}{42} when ± is plus. Add -120 to 90.

$$p=-\frac{30}{42}$$

Reduce the fraction \frac{-30}{42} to lowest terms by extracting and canceling out 6.

$$p=-\frac{5}{7}$$

Now solve the equation p=\frac{-120±90}{42} when ± is minus. Subtract 90 from -120.

$$p=-\frac{210}{42}$$

Divide -210 by 42.

$$p=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{7} for x_{1} and -5 for x_{2}.

$$21p^{2}+120p+75=21\left(p-\left(-\frac{5}{7}\right)\right)\left(p-\left(-5\right)\right)$$

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

$$21p^{2}+120p+75=21\left(p+\frac{5}{7}\right)\left(p+5\right)$$

Add \frac{5}{7} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21p^{2}+120p+75=21\times \frac{7p+5}{7}\left(p+5\right)$$

Cancel out 7, the greatest common factor in 21 and 7.

$$21p^{2}+120p+75=3\left(7p+5\right)\left(p+5\right)$$

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 21

$$x ^ 2 +\frac{40}{7}x +\frac{25}{7} = 0$$

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

$$r + s = -\frac{40}{7} rs = \frac{25}{7}$$

Two numbers r and s sum up to -\frac{40}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{40}{7} = -\frac{20}{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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{20}{7} - u s = -\frac{20}{7} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{7}

$$(-\frac{20}{7} - u) (-\frac{20}{7} + u) = \frac{25}{7}$$

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

$$\frac{400}{49} - u^2 = \frac{25}{7}$$

Simplify the expression by subtracting \frac{400}{49} on both sides

$$-u^2 = \frac{25}{7}-\frac{400}{49} = -\frac{225}{49}$$

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

$$u^2 = \frac{225}{49} u = \pm\sqrt{\frac{225}{49}} = \pm \frac{15}{7}$$

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

$$r =-\frac{20}{7} - \frac{15}{7} = -5 s = -\frac{20}{7} + \frac{15}{7} = -0.714$$

Evaluate

$$3\left(p+5\right)\left(7p+5\right)$$