Solve x^2+4x-21-0

$${ x }^{ 2 } +4x-21-0$$

Evaluate

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

Solution Steps

Multiply -1 and 0 to get 0.

$$x^{2}+4x-21+0$$

Add -21 and 0 to get -21.

$$x^{2}+4x-21$$

Factor

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

Solution Steps

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

$$a+b=4 ab=1\left(-21\right)=-21$$

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 -21.

$$-1,21 -3,7$$

Calculate the sum for each pair.

$$-1+21=20 -3+7=4$$

The solution is the pair that gives sum 4.

$$a=-3 b=7$$

Rewrite x^{2}+4x-21 as \left(x^{2}-3x\right)+\left(7x-21\right).

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

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

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

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

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

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.

$$x^{2}+4x-21=0$$

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.

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

Square 4.

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

Multiply -4 times -21.

$$x=\frac{-4±\sqrt{16+84}}{2}$$

Add 16 to 84.

$$x=\frac{-4±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-4±10}{2}$$

Now solve the equation x=\frac{-4±10}{2} when ± is plus. Add -4 to 10.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-4±10}{2} when ± is minus. Subtract 10 from -4.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

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 -7 for x_{2}.

$$x^{2}+4x-21=\left(x-3\right)\left(x-\left(-7\right)\right)$$

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

$$x^{2}+4x-21=\left(x-3\right)\left(x+7\right)$$