Solve 3x^2-16x+21

$$3 x ^ { 2 } - 16 x + 21$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=-16 ab=3\times 21=63$$

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 63.

$$-1,-63 -3,-21 -7,-9$$

Calculate the sum for each pair.

$$-1-63=-64 -3-21=-24 -7-9=-16$$

The solution is the pair that gives sum -16.

$$a=-9 b=-7$$

Rewrite 3x^{2}-16x+21 as \left(3x^{2}-9x\right)+\left(-7x+21\right).

$$\left(3x^{2}-9x\right)+\left(-7x+21\right)$$

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

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

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

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

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.

$$3x^{2}-16x+21=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 3\times 21}}{2\times 3}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 3\times 21}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-16\right)±\sqrt{256-12\times 21}}{2\times 3}$$

Multiply -12 times 21.

$$x=\frac{-\left(-16\right)±\sqrt{256-252}}{2\times 3}$$

Add 256 to -252.

$$x=\frac{-\left(-16\right)±\sqrt{4}}{2\times 3}$$

Take the square root of 4.

$$x=\frac{-\left(-16\right)±2}{2\times 3}$$

The opposite of -16 is 16.

$$x=\frac{16±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{16±2}{6}$$

Now solve the equation x=\frac{16±2}{6} when ± is plus. Add 16 to 2.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{16±2}{6} when ± is minus. Subtract 2 from 16.

$$x=\frac{14}{6}$$

Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

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

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

$$3x^{2}-16x+21=3\left(x-3\right)\left(x-\frac{7}{3}\right)$$

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

$$3x^{2}-16x+21=3\left(x-3\right)\times \frac{3x-7}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-16x+21=\left(x-3\right)\left(3x-7\right)$$

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 3

$$x ^ 2 -\frac{16}{3}x +7 = 0$$

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

$$r + s = \frac{16}{3} rs = 7$$

Two numbers r and s sum up to \frac{16}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{3} = \frac{8}{3}. 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>

$$r = \frac{8}{3} - u s = \frac{8}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 7

$$(\frac{8}{3} - u) (\frac{8}{3} + u) = 7$$

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

$$\frac{64}{9} - u^2 = 7$$

Simplify the expression by subtracting \frac{64}{9} on both sides

$$-u^2 = 7-\frac{64}{9} = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =\frac{8}{3} - \frac{1}{3} = 2.333 s = \frac{8}{3} + \frac{1}{3} = 3.000$$

Evaluate

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