Solve 15p^2+65p+50

$$15 p ^ { 2 } + 65 p + 50$$

Factor

$$5\left(p+1\right)\left(3p+10\right)$$

Solution Steps

Factor out 5.

$$5\left(3p^{2}+13p+10\right)$$

Consider 3p^{2}+13p+10. Factor the expression by grouping. First, the expression needs to be rewritten as 3p^{2}+ap+bp+10. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\times 10=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=3 b=10$$

Rewrite 3p^{2}+13p+10 as \left(3p^{2}+3p\right)+\left(10p+10\right).

$$\left(3p^{2}+3p\right)+\left(10p+10\right)$$

Factor out 3p in the first and 10 in the second group.

$$3p\left(p+1\right)+10\left(p+1\right)$$

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

$$\left(p+1\right)\left(3p+10\right)$$

Rewrite the complete factored expression.

$$5\left(p+1\right)\left(3p+10\right)$$

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.

$$15p^{2}+65p+50=0$$

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.

$$p=\frac{-65±\sqrt{65^{2}-4\times 15\times 50}}{2\times 15}$$

Square 65.

$$p=\frac{-65±\sqrt{4225-4\times 15\times 50}}{2\times 15}$$

Multiply -4 times 15.

$$p=\frac{-65±\sqrt{4225-60\times 50}}{2\times 15}$$

Multiply -60 times 50.

$$p=\frac{-65±\sqrt{4225-3000}}{2\times 15}$$

Add 4225 to -3000.

$$p=\frac{-65±\sqrt{1225}}{2\times 15}$$

Take the square root of 1225.

$$p=\frac{-65±35}{2\times 15}$$

Multiply 2 times 15.

$$p=\frac{-65±35}{30}$$

Now solve the equation p=\frac{-65±35}{30} when ± is plus. Add -65 to 35.

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

Divide -30 by 30.

$$p=-1$$

Now solve the equation p=\frac{-65±35}{30} when ± is minus. Subtract 35 from -65.

$$p=-\frac{100}{30}$$

Reduce the fraction \frac{-100}{30} to lowest terms by extracting and canceling out 10.

$$p=-\frac{10}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -\frac{10}{3} for x_{2}.

$$15p^{2}+65p+50=15\left(p-\left(-1\right)\right)\left(p-\left(-\frac{10}{3}\right)\right)$$

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

$$15p^{2}+65p+50=15\left(p+1\right)\left(p+\frac{10}{3}\right)$$

Add \frac{10}{3} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15p^{2}+65p+50=15\left(p+1\right)\times \frac{3p+10}{3}$$

Cancel out 3, the greatest common factor in 15 and 3.

$$15p^{2}+65p+50=5\left(p+1\right)\left(3p+10\right)$$

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 15

$$x ^ 2 +\frac{13}{3}x +\frac{10}{3} = 0$$

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

$$r + s = -\frac{13}{3} rs = \frac{10}{3}$$

Two numbers r and s sum up to -\frac{13}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{3} = -\frac{13}{6}. 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>

$$r = -\frac{13}{6} - u s = -\frac{13}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{10}{3}

$$(-\frac{13}{6} - u) (-\frac{13}{6} + u) = \frac{10}{3}$$

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

$$\frac{169}{36} - u^2 = \frac{10}{3}$$

Simplify the expression by subtracting \frac{169}{36} on both sides

$$-u^2 = \frac{10}{3}-\frac{169}{36} = -\frac{49}{36}$$

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

$$u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}$$

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

$$r =-\frac{13}{6} - \frac{7}{6} = -3.333 s = -\frac{13}{6} + \frac{7}{6} = -1.000$$

Evaluate

$$5\left(p+1\right)\left(3p+10\right)$$