Solve q^2-6q-7

$$q ^ { 2 } - 6 q - 7$$

Factor

$$\left(q-7\right)\left(q+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as q^{2}+aq+bq-7. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-7\right)=-7$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-7 b=1$$

Rewrite q^{2}-6q-7 as \left(q^{2}-7q\right)+\left(q-7\right).

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

Factor out q in q^{2}-7q.

$$q\left(q-7\right)+q-7$$

Factor out common term q-7 by using distributive property.

$$\left(q-7\right)\left(q+1\right)$$

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.

$$q^{2}-6q-7=0$$

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.

$$q=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-7\right)}}{2}$$

Square -6.

$$q=\frac{-\left(-6\right)±\sqrt{36-4\left(-7\right)}}{2}$$

Multiply -4 times -7.

$$q=\frac{-\left(-6\right)±\sqrt{36+28}}{2}$$

Add 36 to 28.

$$q=\frac{-\left(-6\right)±\sqrt{64}}{2}$$

Take the square root of 64.

$$q=\frac{-\left(-6\right)±8}{2}$$

The opposite of -6 is 6.

$$q=\frac{6±8}{2}$$

Now solve the equation q=\frac{6±8}{2} when ± is plus. Add 6 to 8.

$$q=\frac{14}{2}$$

Divide 14 by 2.

$$q=7$$

Now solve the equation q=\frac{6±8}{2} when ± is minus. Subtract 8 from 6.

$$q=-\frac{2}{2}$$

Divide -2 by 2.

$$q=-1$$

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

$$q^{2}-6q-7=\left(q-7\right)\left(q-\left(-1\right)\right)$$

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

$$q^{2}-6q-7=\left(q-7\right)\left(q+1\right)$$

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.

$$x ^ 2 -6x -7 = 0$$

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

$$r + s = 6 rs = -7$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -7

$$(3 - u) (3 + u) = -7$$

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

$$9 - u^2 = -7$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = -7-9 = -16$$

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

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

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

$$r =3 - 4 = -1 s = 3 + 4 = 7$$

Evaluate

$$\left(q-7\right)\left(q+1\right)$$