Solve 2a^2+9a+10

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

Factor

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

Solution Steps

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

$$p+q=9 pq=2\times 10=20$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q 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.

$$p=4 q=5$$

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

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

Factor out 2a in the first and 5 in the second group.

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

Factor out common term a+2 by using distributive property.

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

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.

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

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.

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

Square 9.

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

Multiply -4 times 2.

$$a=\frac{-9±\sqrt{81-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

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

Add 81 to -80.

$$a=\frac{-9±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$a=\frac{-9±1}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{-9±1}{4}$$

Now solve the equation a=\frac{-9±1}{4} when ± is plus. Add -9 to 1.

$$a=-\frac{8}{4}$$

Divide -8 by 4.

$$a=-2$$

Now solve the equation a=\frac{-9±1}{4} when ± is minus. Subtract 1 from -9.

$$a=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

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

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

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

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

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

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

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

Cancel out 2, the greatest common factor in 2 and 2.

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

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 2

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

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

$$r + s = -\frac{9}{2} rs = 5$$

Two numbers r and s sum up to -\frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{2} = -\frac{9}{4}. 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>

$$r = -\frac{9}{4} - u s = -\frac{9}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 5

$$(-\frac{9}{4} - u) (-\frac{9}{4} + u) = 5$$

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

$$\frac{81}{16} - u^2 = 5$$

Simplify the expression by subtracting \frac{81}{16} on both sides

$$-u^2 = 5-\frac{81}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =-\frac{9}{4} - \frac{1}{4} = -2.500 s = -\frac{9}{4} + \frac{1}{4} = -2$$

Evaluate

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