Solve 4x^2-41x+45

$$4 x ^ { 2 } - 41 x + 45$$

Factor

$$\left(x-9\right)\left(4x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+45. To find a and b, set up a system to be solved.

$$a+b=-41 ab=4\times 45=180$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 -41.

$$a=-36 b=-5$$

Rewrite 4x^{2}-41x+45 as \left(4x^{2}-36x\right)+\left(-5x+45\right).

$$\left(4x^{2}-36x\right)+\left(-5x+45\right)$$

Factor out 4x in the first and -5 in the second group.

$$4x\left(x-9\right)-5\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(4x-5\right)$$

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.

$$4x^{2}-41x+45=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 4\times 45}}{2\times 4}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 4\times 45}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-41\right)±\sqrt{1681-16\times 45}}{2\times 4}$$

Multiply -16 times 45.

$$x=\frac{-\left(-41\right)±\sqrt{1681-720}}{2\times 4}$$

Add 1681 to -720.

$$x=\frac{-\left(-41\right)±\sqrt{961}}{2\times 4}$$

Take the square root of 961.

$$x=\frac{-\left(-41\right)±31}{2\times 4}$$

The opposite of -41 is 41.

$$x=\frac{41±31}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{41±31}{8}$$

Now solve the equation x=\frac{41±31}{8} when ± is plus. Add 41 to 31.

$$x=\frac{72}{8}$$

Divide 72 by 8.

$$x=9$$

Now solve the equation x=\frac{41±31}{8} when ± is minus. Subtract 31 from 41.

$$x=\frac{10}{8}$$

Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and \frac{5}{4} for x_{2}.

$$4x^{2}-41x+45=4\left(x-9\right)\left(x-\frac{5}{4}\right)$$

Subtract \frac{5}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-41x+45=4\left(x-9\right)\times \frac{4x-5}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4x^{2}-41x+45=\left(x-9\right)\left(4x-5\right)$$

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 4

$$x ^ 2 -\frac{41}{4}x +\frac{45}{4} = 0$$

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

$$r + s = \frac{41}{4} rs = \frac{45}{4}$$

Two numbers r and s sum up to \frac{41}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{41}{4} = \frac{41}{8}. 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>

$$r = \frac{41}{8} - u s = \frac{41}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{45}{4}

$$(\frac{41}{8} - u) (\frac{41}{8} + u) = \frac{45}{4}$$

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

$$\frac{1681}{64} - u^2 = \frac{45}{4}$$

Simplify the expression by subtracting \frac{1681}{64} on both sides

$$-u^2 = \frac{45}{4}-\frac{1681}{64} = -\frac{961}{64}$$

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

$$u^2 = \frac{961}{64} u = \pm\sqrt{\frac{961}{64}} = \pm \frac{31}{8}$$

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

$$r =\frac{41}{8} - \frac{31}{8} = 1.250 s = \frac{41}{8} + \frac{31}{8} = 9$$

Evaluate

$$\left(x-9\right)\left(4x-5\right)$$