Solve 21x^2+10x-16

$$21 x ^ { 2 } + 10 x - 16$$

Factor

$$\left(3x-2\right)\left(7x+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=10 ab=21\left(-16\right)=-336$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -336.

$$-1,336 -2,168 -3,112 -4,84 -6,56 -7,48 -8,42 -12,28 -14,24 -16,21$$

Calculate the sum for each pair.

$$-1+336=335 -2+168=166 -3+112=109 -4+84=80 -6+56=50 -7+48=41 -8+42=34 -12+28=16 -14+24=10 -16+21=5$$

The solution is the pair that gives sum 10.

$$a=-14 b=24$$

Rewrite 21x^{2}+10x-16 as \left(21x^{2}-14x\right)+\left(24x-16\right).

$$\left(21x^{2}-14x\right)+\left(24x-16\right)$$

Factor out 7x in the first and 8 in the second group.

$$7x\left(3x-2\right)+8\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(7x+8\right)$$

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.

$$21x^{2}+10x-16=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 21\left(-16\right)}}{2\times 21}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 21\left(-16\right)}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-10±\sqrt{100-84\left(-16\right)}}{2\times 21}$$

Multiply -84 times -16.

$$x=\frac{-10±\sqrt{100+1344}}{2\times 21}$$

Add 100 to 1344.

$$x=\frac{-10±\sqrt{1444}}{2\times 21}$$

Take the square root of 1444.

$$x=\frac{-10±38}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-10±38}{42}$$

Now solve the equation x=\frac{-10±38}{42} when ± is plus. Add -10 to 38.

$$x=\frac{28}{42}$$

Reduce the fraction \frac{28}{42} to lowest terms by extracting and canceling out 14.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{-10±38}{42} when ± is minus. Subtract 38 from -10.

$$x=-\frac{48}{42}$$

Reduce the fraction \frac{-48}{42} to lowest terms by extracting and canceling out 6.

$$x=-\frac{8}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{3} for x_{1} and -\frac{8}{7} for x_{2}.

$$21x^{2}+10x-16=21\left(x-\frac{2}{3}\right)\left(x-\left(-\frac{8}{7}\right)\right)$$

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

$$21x^{2}+10x-16=21\left(x-\frac{2}{3}\right)\left(x+\frac{8}{7}\right)$$

Subtract \frac{2}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+10x-16=21\times \frac{3x-2}{3}\left(x+\frac{8}{7}\right)$$

Add \frac{8}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+10x-16=21\times \frac{3x-2}{3}\times \frac{7x+8}{7}$$

Multiply \frac{3x-2}{3} times \frac{7x+8}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+10x-16=21\times \frac{\left(3x-2\right)\left(7x+8\right)}{3\times 7}$$

Multiply 3 times 7.

$$21x^{2}+10x-16=21\times \frac{\left(3x-2\right)\left(7x+8\right)}{21}$$

Cancel out 21, the greatest common factor in 21 and 21.

$$21x^{2}+10x-16=\left(3x-2\right)\left(7x+8\right)$$

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 21

$$x ^ 2 +\frac{10}{21}x -\frac{16}{21} = 0$$

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

$$r + s = -\frac{10}{21} rs = -\frac{16}{21}$$

Two numbers r and s sum up to -\frac{10}{21} exactly when the average of the two numbers is \frac{1}{2}*-\frac{10}{21} = -\frac{5}{21}. 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>

$$r = -\frac{5}{21} - u s = -\frac{5}{21} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{16}{21}

$$(-\frac{5}{21} - u) (-\frac{5}{21} + u) = -\frac{16}{21}$$

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

$$\frac{25}{441} - u^2 = -\frac{16}{21}$$

Simplify the expression by subtracting \frac{25}{441} on both sides

$$-u^2 = -\frac{16}{21}-\frac{25}{441} = -\frac{361}{441}$$

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

$$u^2 = \frac{361}{441} u = \pm\sqrt{\frac{361}{441}} = \pm \frac{19}{21}$$

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

$$r =-\frac{5}{21} - \frac{19}{21} = -1.143 s = -\frac{5}{21} + \frac{19}{21} = 0.667$$

Evaluate

$$\left(3x-2\right)\left(7x+8\right)$$