Solve 6x^2+15x-21

$$6 x ^ { 2 } + 15 x - 21$$

Factor

$$3\left(x-1\right)\left(2x+7\right)$$

Solution Steps

Factor out 3.

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

Consider 2x^{2}+5x-7. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=5 ab=2\left(-7\right)=-14$$

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 -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite 2x^{2}+5x-7 as \left(2x^{2}-2x\right)+\left(7x-7\right).

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

Factor out 2x in the first and 7 in the second group.

$$2x\left(x-1\right)+7\left(x-1\right)$$

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

$$\left(x-1\right)\left(2x+7\right)$$

Rewrite the complete factored expression.

$$3\left(x-1\right)\left(2x+7\right)$$

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.

$$6x^{2}+15x-21=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 6\left(-21\right)}}{2\times 6}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 6\left(-21\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-15±\sqrt{225-24\left(-21\right)}}{2\times 6}$$

Multiply -24 times -21.

$$x=\frac{-15±\sqrt{225+504}}{2\times 6}$$

Add 225 to 504.

$$x=\frac{-15±\sqrt{729}}{2\times 6}$$

Take the square root of 729.

$$x=\frac{-15±27}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-15±27}{12}$$

Now solve the equation x=\frac{-15±27}{12} when ± is plus. Add -15 to 27.

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

Divide 12 by 12.

$$x=1$$

Now solve the equation x=\frac{-15±27}{12} when ± is minus. Subtract 27 from -15.

$$x=-\frac{42}{12}$$

Reduce the fraction \frac{-42}{12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{7}{2}$$

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

$$6x^{2}+15x-21=6\left(x-1\right)\left(x-\left(-\frac{7}{2}\right)\right)$$

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

$$6x^{2}+15x-21=6\left(x-1\right)\left(x+\frac{7}{2}\right)$$

Add \frac{7}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+15x-21=6\left(x-1\right)\times \frac{2x+7}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}+15x-21=3\left(x-1\right)\left(2x+7\right)$$

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

$$x ^ 2 +\frac{5}{2}x -\frac{7}{2} = 0$$

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

$$r + s = -\frac{5}{2} rs = -\frac{7}{2}$$

Two numbers r and s sum up to -\frac{5}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{2} = -\frac{5}{4}. 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>

$$r = -\frac{5}{4} - u s = -\frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{2}

$$(-\frac{5}{4} - u) (-\frac{5}{4} + u) = -\frac{7}{2}$$

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

$$\frac{25}{16} - u^2 = -\frac{7}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = -\frac{7}{2}-\frac{25}{16} = -\frac{81}{16}$$

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

$$u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}$$

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

$$r =-\frac{5}{4} - \frac{9}{4} = -3.500 s = -\frac{5}{4} + \frac{9}{4} = 1$$

Evaluate

$$3\left(x-1\right)\left(2x+7\right)$$