Alg7.1.1

From Exampleproblems

$x^2 < 10 - 3x\,$

Move everything to one side. It's good to keep the coefficient of $x 2$ positive. It's better to have it be equal to one.

$x^2+3x-10<0\,$

Factor the polynomial. To do so, find a pair of numbers that will multiply together to give the constant term $- 10$ . $(1,-10),(-1,10),(2,-5),(-2,5)\,$ . But the pair $(-2,5)\,$ will also sum to $3$ , which is the coefficient of $x$ . So the factors are

$x^2+3x-10 = (x-2)(x+5)\,$

Verify this by taking the product of the right hand side in your head. Take as long as you need but don't resort to pencil and paper.

Now, set up the equation (with the 'equals' sign)

$(x-2)(x+5)=0\,$

The roots (critical points) are $x=2,-5\,$ since these values of $x$ will make one of the factors equal zero.

These critical points split up all the numbers into intervals, and the answer will be some combination of these intervals of all numbers below $- 5$ , between $- 5$ and $2$ , and all numbers above $2$ .

According to the original problem, the answer should be the intervals that have values that can be plugged in to $x$ to get numbers less than $0$ . Testing one point in the interval will determine whether all the numbers belong or not. Pick easy numbers to test.

In the interval $(-\infty, -5)\,$ , choose (for example) $- 10$ . $(-10-2)(-10+5)\,$ gives a negative number times a negative number, so all these numbers will be positive. They won't be included.