Solve 21x^2+32x-5

$$21 x ^ { 2 } + 32 x - 5$$

Factor

$$\left(7x-1\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=32 ab=21\left(-5\right)=-105$$

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 -105.

$$-1,105 -3,35 -5,21 -7,15$$

Calculate the sum for each pair.

$$-1+105=104 -3+35=32 -5+21=16 -7+15=8$$

The solution is the pair that gives sum 32.

$$a=-3 b=35$$

Rewrite 21x^{2}+32x-5 as \left(21x^{2}-3x\right)+\left(35x-5\right).

$$\left(21x^{2}-3x\right)+\left(35x-5\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(7x-1\right)+5\left(7x-1\right)$$

Factor out common term 7x-1 by using distributive property.

$$\left(7x-1\right)\left(3x+5\right)$$

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.

$$21x^{2}+32x-5=0$$

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.

$$x=\frac{-32±\sqrt{32^{2}-4\times 21\left(-5\right)}}{2\times 21}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 21\left(-5\right)}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-32±\sqrt{1024-84\left(-5\right)}}{2\times 21}$$

Multiply -84 times -5.

$$x=\frac{-32±\sqrt{1024+420}}{2\times 21}$$

Add 1024 to 420.

$$x=\frac{-32±\sqrt{1444}}{2\times 21}$$

Take the square root of 1444.

$$x=\frac{-32±38}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-32±38}{42}$$

Now solve the equation x=\frac{-32±38}{42} when ± is plus. Add -32 to 38.

$$x=\frac{6}{42}$$

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

$$x=\frac{1}{7}$$

Now solve the equation x=\frac{-32±38}{42} when ± is minus. Subtract 38 from -32.

$$x=-\frac{70}{42}$$

Reduce the fraction \frac{-70}{42} to lowest terms by extracting and canceling out 14.

$$x=-\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{7} for x_{1} and -\frac{5}{3} for x_{2}.

$$21x^{2}+32x-5=21\left(x-\frac{1}{7}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$21x^{2}+32x-5=21\left(x-\frac{1}{7}\right)\left(x+\frac{5}{3}\right)$$

Subtract \frac{1}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+32x-5=21\times \frac{7x-1}{7}\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+32x-5=21\times \frac{7x-1}{7}\times \frac{3x+5}{3}$$

Multiply \frac{7x-1}{7} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21x^{2}+32x-5=21\times \frac{\left(7x-1\right)\left(3x+5\right)}{7\times 3}$$

Multiply 7 times 3.

$$21x^{2}+32x-5=21\times \frac{\left(7x-1\right)\left(3x+5\right)}{21}$$

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

$$21x^{2}+32x-5=\left(7x-1\right)\left(3x+5\right)$$

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 21

$$x ^ 2 +\frac{32}{21}x -\frac{5}{21} = 0$$

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

$$r + s = -\frac{32}{21} rs = -\frac{5}{21}$$

Two numbers r and s sum up to -\frac{32}{21} exactly when the average of the two numbers is \frac{1}{2}*-\frac{32}{21} = -\frac{16}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{21}

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

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

$$\frac{256}{441} - u^2 = -\frac{5}{21}$$

Simplify the expression by subtracting \frac{256}{441} on both sides

$$-u^2 = -\frac{5}{21}-\frac{256}{441} = -\frac{361}{441}$$

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.

$$r =-\frac{16}{21} - \frac{19}{21} = -1.667 s = -\frac{16}{21} + \frac{19}{21} = 0.143$$

Evaluate

$$\left(7x-1\right)\left(3x+5\right)$$