Solve 3p^2+34p+80

$$3 p ^ { 2 } + 34 p + 80$$

Factor

$$\left(p+8\right)\left(3p+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3p^{2}+ap+bp+80. To find a and b, set up a system to be solved.

$$a+b=34 ab=3\times 80=240$$

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 240.

$$1,240 2,120 3,80 4,60 5,48 6,40 8,30 10,24 12,20 15,16$$

Calculate the sum for each pair.

$$1+240=241 2+120=122 3+80=83 4+60=64 5+48=53 6+40=46 8+30=38 10+24=34 12+20=32 15+16=31$$

The solution is the pair that gives sum 34.

$$a=10 b=24$$

Rewrite 3p^{2}+34p+80 as \left(3p^{2}+10p\right)+\left(24p+80\right).

$$\left(3p^{2}+10p\right)+\left(24p+80\right)$$

Factor out p in the first and 8 in the second group.

$$p\left(3p+10\right)+8\left(3p+10\right)$$

Factor out common term 3p+10 by using distributive property.

$$\left(3p+10\right)\left(p+8\right)$$

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.

$$3p^{2}+34p+80=0$$

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.

$$p=\frac{-34±\sqrt{34^{2}-4\times 3\times 80}}{2\times 3}$$

Square 34.

$$p=\frac{-34±\sqrt{1156-4\times 3\times 80}}{2\times 3}$$

Multiply -4 times 3.

$$p=\frac{-34±\sqrt{1156-12\times 80}}{2\times 3}$$

Multiply -12 times 80.

$$p=\frac{-34±\sqrt{1156-960}}{2\times 3}$$

Add 1156 to -960.

$$p=\frac{-34±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$p=\frac{-34±14}{2\times 3}$$

Multiply 2 times 3.

$$p=\frac{-34±14}{6}$$

Now solve the equation p=\frac{-34±14}{6} when ± is plus. Add -34 to 14.

$$p=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation p=\frac{-34±14}{6} when ± is minus. Subtract 14 from -34.

$$p=-\frac{48}{6}$$

Divide -48 by 6.

$$p=-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{10}{3} for x_{1} and -8 for x_{2}.

$$3p^{2}+34p+80=3\left(p-\left(-\frac{10}{3}\right)\right)\left(p-\left(-8\right)\right)$$

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

$$3p^{2}+34p+80=3\left(p+\frac{10}{3}\right)\left(p+8\right)$$

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

$$3p^{2}+34p+80=3\times \frac{3p+10}{3}\left(p+8\right)$$

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

$$3p^{2}+34p+80=\left(3p+10\right)\left(p+8\right)$$

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 3

$$x ^ 2 +\frac{34}{3}x +\frac{80}{3} = 0$$

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

$$r + s = -\frac{34}{3} rs = \frac{80}{3}$$

Two numbers r and s sum up to -\frac{34}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{34}{3} = -\frac{17}{3}. 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>

$$r = -\frac{17}{3} - u s = -\frac{17}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{80}{3}

$$(-\frac{17}{3} - u) (-\frac{17}{3} + u) = \frac{80}{3}$$

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

$$\frac{289}{9} - u^2 = \frac{80}{3}$$

Simplify the expression by subtracting \frac{289}{9} on both sides

$$-u^2 = \frac{80}{3}-\frac{289}{9} = -\frac{49}{9}$$

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

$$u^2 = \frac{49}{9} u = \pm\sqrt{\frac{49}{9}} = \pm \frac{7}{3}$$

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

$$r =-\frac{17}{3} - \frac{7}{3} = -8 s = -\frac{17}{3} + \frac{7}{3} = -3.333$$

Evaluate

$$\left(p+8\right)\left(3p+10\right)$$