Solve 11x^2-54x+63

$$11 x ^ { 2 } - 54 x + 63$$

Factor

$$\left(x-3\right)\left(11x-21\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 11x^{2}+ax+bx+63. To find a and b, set up a system to be solved.

$$a+b=-54 ab=11\times 63=693$$

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 693.

$$-1,-693 -3,-231 -7,-99 -9,-77 -11,-63 -21,-33$$

Calculate the sum for each pair.

$$-1-693=-694 -3-231=-234 -7-99=-106 -9-77=-86 -11-63=-74 -21-33=-54$$

The solution is the pair that gives sum -54.

$$a=-33 b=-21$$

Rewrite 11x^{2}-54x+63 as \left(11x^{2}-33x\right)+\left(-21x+63\right).

$$\left(11x^{2}-33x\right)+\left(-21x+63\right)$$

Factor out 11x in the first and -21 in the second group.

$$11x\left(x-3\right)-21\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(11x-21\right)$$

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.

$$11x^{2}-54x+63=0$$

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.

$$x=\frac{-\left(-54\right)±\sqrt{\left(-54\right)^{2}-4\times 11\times 63}}{2\times 11}$$

Square -54.

$$x=\frac{-\left(-54\right)±\sqrt{2916-4\times 11\times 63}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-\left(-54\right)±\sqrt{2916-44\times 63}}{2\times 11}$$

Multiply -44 times 63.

$$x=\frac{-\left(-54\right)±\sqrt{2916-2772}}{2\times 11}$$

Add 2916 to -2772.

$$x=\frac{-\left(-54\right)±\sqrt{144}}{2\times 11}$$

Take the square root of 144.

$$x=\frac{-\left(-54\right)±12}{2\times 11}$$

The opposite of -54 is 54.

$$x=\frac{54±12}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{54±12}{22}$$

Now solve the equation x=\frac{54±12}{22} when ± is plus. Add 54 to 12.

$$x=\frac{66}{22}$$

Divide 66 by 22.

$$x=3$$

Now solve the equation x=\frac{54±12}{22} when ± is minus. Subtract 12 from 54.

$$x=\frac{42}{22}$$

Reduce the fraction \frac{42}{22} to lowest terms by extracting and canceling out 2.

$$x=\frac{21}{11}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and \frac{21}{11} for x_{2}.

$$11x^{2}-54x+63=11\left(x-3\right)\left(x-\frac{21}{11}\right)$$

Subtract \frac{21}{11} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$11x^{2}-54x+63=11\left(x-3\right)\times \frac{11x-21}{11}$$

Cancel out 11, the greatest common factor in 11 and 11.

$$11x^{2}-54x+63=\left(x-3\right)\left(11x-21\right)$$

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 11

$$x ^ 2 -\frac{54}{11}x +\frac{63}{11} = 0$$

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

$$r + s = \frac{54}{11} rs = \frac{63}{11}$$

Two numbers r and s sum up to \frac{54}{11} exactly when the average of the two numbers is \frac{1}{2}*\frac{54}{11} = \frac{27}{11}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{27}{11} - u s = \frac{27}{11} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{63}{11}

$$(\frac{27}{11} - u) (\frac{27}{11} + u) = \frac{63}{11}$$

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

$$\frac{729}{121} - u^2 = \frac{63}{11}$$

Simplify the expression by subtracting \frac{729}{121} on both sides

$$-u^2 = \frac{63}{11}-\frac{729}{121} = -\frac{36}{121}$$

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

$$u^2 = \frac{36}{121} u = \pm\sqrt{\frac{36}{121}} = \pm \frac{6}{11}$$

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

$$r =\frac{27}{11} - \frac{6}{11} = 1.909 s = \frac{27}{11} + \frac{6}{11} = 3$$

Evaluate

$$\left(x-3\right)\left(11x-21\right)$$