Solve 3b^2+12b-63

$$3 b ^ { 2 } + 12 b - 63$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(b^{2}+4b-21\right)$$

Consider b^{2}+4b-21. Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-21. To find p and q, set up a system to be solved.

$$p+q=4 pq=1\left(-21\right)=-21$$

Since pq is negative, p and q have the opposite signs. Since p+q 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.

$$p=-3 q=7$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$3b^{2}+12b-63=0$$

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.

$$b=\frac{-12±\sqrt{12^{2}-4\times 3\left(-63\right)}}{2\times 3}$$

Square 12.

$$b=\frac{-12±\sqrt{144-4\times 3\left(-63\right)}}{2\times 3}$$

Multiply -4 times 3.

$$b=\frac{-12±\sqrt{144-12\left(-63\right)}}{2\times 3}$$

Multiply -12 times -63.

$$b=\frac{-12±\sqrt{144+756}}{2\times 3}$$

Add 144 to 756.

$$b=\frac{-12±\sqrt{900}}{2\times 3}$$

Take the square root of 900.

$$b=\frac{-12±30}{2\times 3}$$

Multiply 2 times 3.

$$b=\frac{-12±30}{6}$$

Now solve the equation b=\frac{-12±30}{6} when ± is plus. Add -12 to 30.

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

Divide 18 by 6.

$$b=3$$

Now solve the equation b=\frac{-12±30}{6} when ± is minus. Subtract 30 from -12.

$$b=-\frac{42}{6}$$

Divide -42 by 6.

$$b=-7$$

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}.

$$3b^{2}+12b-63=3\left(b-3\right)\left(b-\left(-7\right)\right)$$

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

$$3b^{2}+12b-63=3\left(b-3\right)\left(b+7\right)$$

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

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

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

$$r + s = -4 rs = -21$$

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

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -21

$$(-2 - u) (-2 + u) = -21$$

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

$$4 - u^2 = -21$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -21-4 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =-2 - 5 = -7 s = -2 + 5 = 3$$

Evaluate

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