Solve 6x^2-41x+63

$$6 x ^ { 2 } - 41 x + 63$$

Factor

$$\left(2x-9\right)\left(3x-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+63. To find a and b, set up a system to be solved.

$$a+b=-41 ab=6\times 63=378$$

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 378.

$$-1,-378 -2,-189 -3,-126 -6,-63 -7,-54 -9,-42 -14,-27 -18,-21$$

Calculate the sum for each pair.

$$-1-378=-379 -2-189=-191 -3-126=-129 -6-63=-69 -7-54=-61 -9-42=-51 -14-27=-41 -18-21=-39$$

The solution is the pair that gives sum -41.

$$a=-27 b=-14$$

Rewrite 6x^{2}-41x+63 as \left(6x^{2}-27x\right)+\left(-14x+63\right).

$$\left(6x^{2}-27x\right)+\left(-14x+63\right)$$

Factor out 3x in the first and -7 in the second group.

$$3x\left(2x-9\right)-7\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(3x-7\right)$$

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.

$$6x^{2}-41x+63=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 6\times 63}}{2\times 6}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 6\times 63}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-41\right)±\sqrt{1681-24\times 63}}{2\times 6}$$

Multiply -24 times 63.

$$x=\frac{-\left(-41\right)±\sqrt{1681-1512}}{2\times 6}$$

Add 1681 to -1512.

$$x=\frac{-\left(-41\right)±\sqrt{169}}{2\times 6}$$

Take the square root of 169.

$$x=\frac{-\left(-41\right)±13}{2\times 6}$$

The opposite of -41 is 41.

$$x=\frac{41±13}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{41±13}{12}$$

Now solve the equation x=\frac{41±13}{12} when ± is plus. Add 41 to 13.

$$x=\frac{54}{12}$$

Reduce the fraction \frac{54}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{41±13}{12} when ± is minus. Subtract 13 from 41.

$$x=\frac{28}{12}$$

Reduce the fraction \frac{28}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{2} for x_{1} and \frac{7}{3} for x_{2}.

$$6x^{2}-41x+63=6\left(x-\frac{9}{2}\right)\left(x-\frac{7}{3}\right)$$

Subtract \frac{9}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-41x+63=6\times \frac{2x-9}{2}\left(x-\frac{7}{3}\right)$$

Subtract \frac{7}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-41x+63=6\times \frac{2x-9}{2}\times \frac{3x-7}{3}$$

Multiply \frac{2x-9}{2} times \frac{3x-7}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-41x+63=6\times \frac{\left(2x-9\right)\left(3x-7\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-41x+63=6\times \frac{\left(2x-9\right)\left(3x-7\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}-41x+63=\left(2x-9\right)\left(3x-7\right)$$

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 6

$$x ^ 2 -\frac{41}{6}x +\frac{21}{2} = 0$$

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

$$r + s = \frac{41}{6} rs = \frac{21}{2}$$

Two numbers r and s sum up to \frac{41}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{41}{6} = \frac{41}{12}. 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>

$$r = \frac{41}{12} - u s = \frac{41}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{21}{2}

$$(\frac{41}{12} - u) (\frac{41}{12} + u) = \frac{21}{2}$$

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

$$\frac{1681}{144} - u^2 = \frac{21}{2}$$

Simplify the expression by subtracting \frac{1681}{144} on both sides

$$-u^2 = \frac{21}{2}-\frac{1681}{144} = -\frac{169}{144}$$

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

$$u^2 = \frac{169}{144} u = \pm\sqrt{\frac{169}{144}} = \pm \frac{13}{12}$$

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

$$r =\frac{41}{12} - \frac{13}{12} = 2.333 s = \frac{41}{12} + \frac{13}{12} = 4.500$$

Evaluate

$$\left(2x-9\right)\left(3x-7\right)$$