Solve 10x^2+41x+40

$$10 x ^ { 2 } + 41 x + 40$$

Factor

$$\left(2x+5\right)\left(5x+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=41 ab=10\times 40=400$$

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 400.

$$1,400 2,200 4,100 5,80 8,50 10,40 16,25 20,20$$

Calculate the sum for each pair.

$$1+400=401 2+200=202 4+100=104 5+80=85 8+50=58 10+40=50 16+25=41 20+20=40$$

The solution is the pair that gives sum 41.

$$a=16 b=25$$

Rewrite 10x^{2}+41x+40 as \left(10x^{2}+16x\right)+\left(25x+40\right).

$$\left(10x^{2}+16x\right)+\left(25x+40\right)$$

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

$$2x\left(5x+8\right)+5\left(5x+8\right)$$

Factor out common term 5x+8 by using distributive property.

$$\left(5x+8\right)\left(2x+5\right)$$

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.

$$10x^{2}+41x+40=0$$

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.

$$x=\frac{-41±\sqrt{41^{2}-4\times 10\times 40}}{2\times 10}$$

Square 41.

$$x=\frac{-41±\sqrt{1681-4\times 10\times 40}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-41±\sqrt{1681-40\times 40}}{2\times 10}$$

Multiply -40 times 40.

$$x=\frac{-41±\sqrt{1681-1600}}{2\times 10}$$

Add 1681 to -1600.

$$x=\frac{-41±\sqrt{81}}{2\times 10}$$

Take the square root of 81.

$$x=\frac{-41±9}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-41±9}{20}$$

Now solve the equation x=\frac{-41±9}{20} when ± is plus. Add -41 to 9.

$$x=-\frac{32}{20}$$

Reduce the fraction \frac{-32}{20} to lowest terms by extracting and canceling out 4.

$$x=-\frac{8}{5}$$

Now solve the equation x=\frac{-41±9}{20} when ± is minus. Subtract 9 from -41.

$$x=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{8}{5} for x_{1} and -\frac{5}{2} for x_{2}.

$$10x^{2}+41x+40=10\left(x-\left(-\frac{8}{5}\right)\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$10x^{2}+41x+40=10\left(x+\frac{8}{5}\right)\left(x+\frac{5}{2}\right)$$

Add \frac{8}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+41x+40=10\times \frac{5x+8}{5}\left(x+\frac{5}{2}\right)$$

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

$$10x^{2}+41x+40=10\times \frac{5x+8}{5}\times \frac{2x+5}{2}$$

Multiply \frac{5x+8}{5} times \frac{2x+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+41x+40=10\times \frac{\left(5x+8\right)\left(2x+5\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}+41x+40=10\times \frac{\left(5x+8\right)\left(2x+5\right)}{10}$$

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

$$10x^{2}+41x+40=\left(5x+8\right)\left(2x+5\right)$$

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 10

$$x ^ 2 +\frac{41}{10}x +4 = 0$$

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

$$r + s = -\frac{41}{10} rs = 4$$

Two numbers r and s sum up to -\frac{41}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{41}{10} = -\frac{41}{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>

$$r = -\frac{41}{20} - u s = -\frac{41}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 4

$$(-\frac{41}{20} - u) (-\frac{41}{20} + u) = 4$$

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

$$\frac{1681}{400} - u^2 = 4$$

Simplify the expression by subtracting \frac{1681}{400} on both sides

$$-u^2 = 4-\frac{1681}{400} = -\frac{81}{400}$$

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

$$u^2 = \frac{81}{400} u = \pm\sqrt{\frac{81}{400}} = \pm \frac{9}{20}$$

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

$$r =-\frac{41}{20} - \frac{9}{20} = -2.500 s = -\frac{41}{20} + \frac{9}{20} = -1.600$$

Evaluate

$$\left(2x+5\right)\left(5x+8\right)$$