Solve 40a^2+126a+44

$$40 a ^ { 2 } + 126 a + 44$$

Factor

$$2\left(4a+11\right)\left(5a+2\right)$$

Solution Steps

Factor out 2.

$$2\left(20a^{2}+63a+22\right)$$

Consider 20a^{2}+63a+22. Factor the expression by grouping. First, the expression needs to be rewritten as 20a^{2}+pa+qa+22. To find p and q, set up a system to be solved.

$$p+q=63 pq=20\times 22=440$$

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 440.

$$1,440 2,220 4,110 5,88 8,55 10,44 11,40 20,22$$

Calculate the sum for each pair.

$$1+440=441 2+220=222 4+110=114 5+88=93 8+55=63 10+44=54 11+40=51 20+22=42$$

The solution is the pair that gives sum 63.

$$p=8 q=55$$

Rewrite 20a^{2}+63a+22 as \left(20a^{2}+8a\right)+\left(55a+22\right).

$$\left(20a^{2}+8a\right)+\left(55a+22\right)$$

Factor out 4a in the first and 11 in the second group.

$$4a\left(5a+2\right)+11\left(5a+2\right)$$

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

$$\left(5a+2\right)\left(4a+11\right)$$

Rewrite the complete factored expression.

$$2\left(5a+2\right)\left(4a+11\right)$$

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.

$$40a^{2}+126a+44=0$$

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.

$$a=\frac{-126±\sqrt{126^{2}-4\times 40\times 44}}{2\times 40}$$

Square 126.

$$a=\frac{-126±\sqrt{15876-4\times 40\times 44}}{2\times 40}$$

Multiply -4 times 40.

$$a=\frac{-126±\sqrt{15876-160\times 44}}{2\times 40}$$

Multiply -160 times 44.

$$a=\frac{-126±\sqrt{15876-7040}}{2\times 40}$$

Add 15876 to -7040.

$$a=\frac{-126±\sqrt{8836}}{2\times 40}$$

Take the square root of 8836.

$$a=\frac{-126±94}{2\times 40}$$

Multiply 2 times 40.

$$a=\frac{-126±94}{80}$$

Now solve the equation a=\frac{-126±94}{80} when ± is plus. Add -126 to 94.

$$a=-\frac{32}{80}$$

Reduce the fraction \frac{-32}{80} to lowest terms by extracting and canceling out 16.

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

Now solve the equation a=\frac{-126±94}{80} when ± is minus. Subtract 94 from -126.

$$a=-\frac{220}{80}$$

Reduce the fraction \frac{-220}{80} to lowest terms by extracting and canceling out 20.

$$a=-\frac{11}{4}$$

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{11}{4} for x_{2}.

$$40a^{2}+126a+44=40\left(a-\left(-\frac{2}{5}\right)\right)\left(a-\left(-\frac{11}{4}\right)\right)$$

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

$$40a^{2}+126a+44=40\left(a+\frac{2}{5}\right)\left(a+\frac{11}{4}\right)$$

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

$$40a^{2}+126a+44=40\times \frac{5a+2}{5}\left(a+\frac{11}{4}\right)$$

Add \frac{11}{4} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$40a^{2}+126a+44=40\times \frac{5a+2}{5}\times \frac{4a+11}{4}$$

Multiply \frac{5a+2}{5} times \frac{4a+11}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$40a^{2}+126a+44=40\times \frac{\left(5a+2\right)\left(4a+11\right)}{5\times 4}$$

Multiply 5 times 4.

$$40a^{2}+126a+44=40\times \frac{\left(5a+2\right)\left(4a+11\right)}{20}$$

Cancel out 20, the greatest common factor in 40 and 20.

$$40a^{2}+126a+44=2\left(5a+2\right)\left(4a+11\right)$$

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 40

$$x ^ 2 +\frac{63}{20}x +\frac{11}{10} = 0$$

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

$$r + s = -\frac{63}{20} rs = \frac{11}{10}$$

Two numbers r and s sum up to -\frac{63}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{63}{20} = -\frac{63}{40}. 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>

$$r = -\frac{63}{40} - u s = -\frac{63}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{11}{10}

$$(-\frac{63}{40} - u) (-\frac{63}{40} + u) = \frac{11}{10}$$

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

$$\frac{3969}{1600} - u^2 = \frac{11}{10}$$

Simplify the expression by subtracting \frac{3969}{1600} on both sides

$$-u^2 = \frac{11}{10}-\frac{3969}{1600} = -\frac{2209}{1600}$$

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

$$u^2 = \frac{2209}{1600} u = \pm\sqrt{\frac{2209}{1600}} = \pm \frac{47}{40}$$

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

$$r =-\frac{63}{40} - \frac{47}{40} = -2.750 s = -\frac{63}{40} + \frac{47}{40} = -0.400$$

Evaluate

$$40a^{2}+126a+44$$