Solve 10p^2+9p+2

$$10 p ^ { 2 } + 9 p + 2$$

Factor

$$\left(2p+1\right)\left(5p+2\right)$$

Solution Steps

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

$$a+b=9 ab=10\times 2=20$$

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 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 9.

$$a=4 b=5$$

Rewrite 10p^{2}+9p+2 as \left(10p^{2}+4p\right)+\left(5p+2\right).

$$\left(10p^{2}+4p\right)+\left(5p+2\right)$$

Factor out 2p in 10p^{2}+4p.

$$2p\left(5p+2\right)+5p+2$$

Factor out common term 5p+2 by using distributive property.

$$\left(5p+2\right)\left(2p+1\right)$$

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.

$$10p^{2}+9p+2=0$$

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.

$$p=\frac{-9±\sqrt{9^{2}-4\times 10\times 2}}{2\times 10}$$

Square 9.

$$p=\frac{-9±\sqrt{81-4\times 10\times 2}}{2\times 10}$$

Multiply -4 times 10.

$$p=\frac{-9±\sqrt{81-40\times 2}}{2\times 10}$$

Multiply -40 times 2.

$$p=\frac{-9±\sqrt{81-80}}{2\times 10}$$

Add 81 to -80.

$$p=\frac{-9±\sqrt{1}}{2\times 10}$$

Take the square root of 1.

$$p=\frac{-9±1}{2\times 10}$$

Multiply 2 times 10.

$$p=\frac{-9±1}{20}$$

Now solve the equation p=\frac{-9±1}{20} when ± is plus. Add -9 to 1.

$$p=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

$$p=-\frac{2}{5}$$

Now solve the equation p=\frac{-9±1}{20} when ± is minus. Subtract 1 from -9.

$$p=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

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

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

$$10p^{2}+9p+2=10\left(p-\left(-\frac{2}{5}\right)\right)\left(p-\left(-\frac{1}{2}\right)\right)$$

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

$$10p^{2}+9p+2=10\left(p+\frac{2}{5}\right)\left(p+\frac{1}{2}\right)$$

Add \frac{2}{5} to p by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10p^{2}+9p+2=10\times \frac{5p+2}{5}\left(p+\frac{1}{2}\right)$$

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

$$10p^{2}+9p+2=10\times \frac{5p+2}{5}\times \frac{2p+1}{2}$$

Multiply \frac{5p+2}{5} times \frac{2p+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10p^{2}+9p+2=10\times \frac{\left(5p+2\right)\left(2p+1\right)}{5\times 2}$$

Multiply 5 times 2.

$$10p^{2}+9p+2=10\times \frac{\left(5p+2\right)\left(2p+1\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10p^{2}+9p+2=\left(5p+2\right)\left(2p+1\right)$$

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 10

$$x ^ 2 +\frac{9}{10}x +\frac{1}{5} = 0$$

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

$$r + s = -\frac{9}{10} rs = \frac{1}{5}$$

Two numbers r and s sum up to -\frac{9}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{10} = -\frac{9}{20}. 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>

$$r = -\frac{9}{20} - u s = -\frac{9}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}

$$(-\frac{9}{20} - u) (-\frac{9}{20} + u) = \frac{1}{5}$$

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

$$\frac{81}{400} - u^2 = \frac{1}{5}$$

Simplify the expression by subtracting \frac{81}{400} on both sides

$$-u^2 = \frac{1}{5}-\frac{81}{400} = -\frac{1}{400}$$

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

$$u^2 = \frac{1}{400} u = \pm\sqrt{\frac{1}{400}} = \pm \frac{1}{20}$$

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

$$r =-\frac{9}{20} - \frac{1}{20} = -0.500 s = -\frac{9}{20} + \frac{1}{20} = -0.400$$

Evaluate

$$\left(2p+1\right)\left(5p+2\right)$$