Solve q^2-10q+21

$$q ^ { 2 } - 10 q + 21$$

Factor

$$\left(q-7\right)\left(q-3\right)$$

Solution Steps

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

$$a+b=-10 ab=1\times 21=21$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.

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

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-7 b=-3$$

Rewrite q^{2}-10q+21 as \left(q^{2}-7q\right)+\left(-3q+21\right).

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

Factor out q in the first and -3 in the second group.

$$q\left(q-7\right)-3\left(q-7\right)$$

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

$$\left(q-7\right)\left(q-3\right)$$

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.

$$q^{2}-10q+21=0$$

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.

$$q=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 21}}{2}$$

Square -10.

$$q=\frac{-\left(-10\right)±\sqrt{100-4\times 21}}{2}$$

Multiply -4 times 21.

$$q=\frac{-\left(-10\right)±\sqrt{100-84}}{2}$$

Add 100 to -84.

$$q=\frac{-\left(-10\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$q=\frac{-\left(-10\right)±4}{2}$$

The opposite of -10 is 10.

$$q=\frac{10±4}{2}$$

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

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

Divide 14 by 2.

$$q=7$$

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

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

Divide 6 by 2.

$$q=3$$

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 3 for x_{2}.

$$q^{2}-10q+21=\left(q-7\right)\left(q-3\right)$$

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.

$$x ^ 2 -10x +21 = 0$$

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

$$r + s = 10 rs = 21$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>

$$r = 5 - u s = 5 + u$$

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

$$(5 - u) (5 + u) = 21$$

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

$$25 - u^2 = 21$$

Simplify the expression by subtracting 25 on both sides

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

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

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

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

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

Evaluate

$$\left(q-7\right)\left(q-3\right)$$