Solve 20a^2+98a+36

$$20 a ^ { 2 } + 98 a + 36$$

Factor

$$2\left(2a+9\right)\left(5a+2\right)$$

Solution Steps

Factor out 2.

$$2\left(10a^{2}+49a+18\right)$$

Consider 10a^{2}+49a+18. Factor the expression by grouping. First, the expression needs to be rewritten as 10a^{2}+pa+qa+18. To find p and q, set up a system to be solved.

$$p+q=49 pq=10\times 18=180$$

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 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 49.

$$p=4 q=45$$

Rewrite 10a^{2}+49a+18 as \left(10a^{2}+4a\right)+\left(45a+18\right).

$$\left(10a^{2}+4a\right)+\left(45a+18\right)$$

Factor out 2a in the first and 9 in the second group.

$$2a\left(5a+2\right)+9\left(5a+2\right)$$

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

$$\left(5a+2\right)\left(2a+9\right)$$

Rewrite the complete factored expression.

$$2\left(5a+2\right)\left(2a+9\right)$$

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.

$$20a^{2}+98a+36=0$$

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.

$$a=\frac{-98±\sqrt{98^{2}-4\times 20\times 36}}{2\times 20}$$

Square 98.

$$a=\frac{-98±\sqrt{9604-4\times 20\times 36}}{2\times 20}$$

Multiply -4 times 20.

$$a=\frac{-98±\sqrt{9604-80\times 36}}{2\times 20}$$

Multiply -80 times 36.

$$a=\frac{-98±\sqrt{9604-2880}}{2\times 20}$$

Add 9604 to -2880.

$$a=\frac{-98±\sqrt{6724}}{2\times 20}$$

Take the square root of 6724.

$$a=\frac{-98±82}{2\times 20}$$

Multiply 2 times 20.

$$a=\frac{-98±82}{40}$$

Now solve the equation a=\frac{-98±82}{40} when ± is plus. Add -98 to 82.

$$a=-\frac{16}{40}$$

Reduce the fraction \frac{-16}{40} to lowest terms by extracting and canceling out 8.

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

Now solve the equation a=\frac{-98±82}{40} when ± is minus. Subtract 82 from -98.

$$a=-\frac{180}{40}$$

Reduce the fraction \frac{-180}{40} to lowest terms by extracting and canceling out 20.

$$a=-\frac{9}{2}$$

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{9}{2} for x_{2}.

$$20a^{2}+98a+36=20\left(a-\left(-\frac{2}{5}\right)\right)\left(a-\left(-\frac{9}{2}\right)\right)$$

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

$$20a^{2}+98a+36=20\left(a+\frac{2}{5}\right)\left(a+\frac{9}{2}\right)$$

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

$$20a^{2}+98a+36=20\times \frac{5a+2}{5}\left(a+\frac{9}{2}\right)$$

Add \frac{9}{2} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20a^{2}+98a+36=20\times \frac{5a+2}{5}\times \frac{2a+9}{2}$$

Multiply \frac{5a+2}{5} times \frac{2a+9}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$20a^{2}+98a+36=20\times \frac{\left(5a+2\right)\left(2a+9\right)}{5\times 2}$$

Multiply 5 times 2.

$$20a^{2}+98a+36=20\times \frac{\left(5a+2\right)\left(2a+9\right)}{10}$$

Cancel out 10, the greatest common factor in 20 and 10.

$$20a^{2}+98a+36=2\left(5a+2\right)\left(2a+9\right)$$

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 20

$$x ^ 2 +\frac{49}{10}x +\frac{9}{5} = 0$$

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

$$r + s = -\frac{49}{10} rs = \frac{9}{5}$$

Two numbers r and s sum up to -\frac{49}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{49}{10} = -\frac{49}{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>

$$r = -\frac{49}{20} - u s = -\frac{49}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{5}

$$(-\frac{49}{20} - u) (-\frac{49}{20} + u) = \frac{9}{5}$$

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

$$\frac{2401}{400} - u^2 = \frac{9}{5}$$

Simplify the expression by subtracting \frac{2401}{400} on both sides

$$-u^2 = \frac{9}{5}-\frac{2401}{400} = -\frac{1681}{400}$$

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

$$u^2 = \frac{1681}{400} u = \pm\sqrt{\frac{1681}{400}} = \pm \frac{41}{20}$$

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

$$r =-\frac{49}{20} - \frac{41}{20} = -4.500 s = -\frac{49}{20} + \frac{41}{20} = -0.400$$

Evaluate

$$20a^{2}+98a+36$$