How to solve quadratic equations with coefficients greater than 1

Solve factorable quadratic equations (leading coefficient is other than 1)
Description
Exercise Name:	Solve factorable quadratic equations (leading coefficient is other than 1)
Math Missions:	Algebra basics Math Mission, Mathematics I Math Mission, Algebra I Math Mission, Mathematics II Math Mission, Algebra II Math Mission, Precalculus Math Mission, Mathematics III Math Mission
Types of Problems:	1

The Solve factorable quadratic equations (leading coefficient is other than 1) exercise appears under the Algebra basics Math Mission, Mathematics I Math Mission, Algebra I Math Mission, Mathematics II Math Mission, Algebra II Math Mission, Precalculus Math Mission and Mathematics III Math Mission. This exercise practices solving equations via factoring with general quadratics.

Types of Problems

There is one type of problem in this exercise:

1. Solve the quadratic equation: This problem provides a quadratic equation that can be factored over integers. The user is expected to factor the equation and use the factored form, with the Zero Product Property, to solve the equation.
   

   Solve the quadratic

Strategies

Knowledge of the Zero Product Property and experience with factoring techniques will ensure accuracy and efficiency on this exercise.

1. The Zero Product property states that if , then either or is zero.
2. Many of these problems require a greatest common factor to be pulled out before the quadratic is to be factored.
3. The quadratic formula can also be used to ensure accuracy, but efficiency would be challenging with this method.
4. These quadratics have a leading coefficient that is not one, so the factoring is more involved than with monic polynomials

Real-life Applications

1. Quadratics are used to model gravity in physics.
2. Quadratics have several applications in business ranging from cost, revenue, and profit to supply and demand.
3. Knowledge of algebra is essential for higher math levels like trigonometry and calculus. Algebra also has countless applications in the real world.

I am not sure quite where you have encountered this phrase, but I suspect it may be in the context of an explanation as to how you might attempt to factor a quadratic.

When written in standard form, a polynomial is a sum of terms in descending order of power (a.k.a. degree) of #x# (or whatever variable you are using).

For example, the cubic polynomial:

#x^3+5x^2-7#

is in standard form since the degrees of the terms #3, 2, 0# are in descending order.

When written in this way, the "leading" term is the term of highest degree and the "leading coefficient" is the multiplier (coefficient) of this term.

In the case of #x^3+5x^2-7#, the "leading coefficient" is #1#, because we could have equivalently written #1x^3#.

A polynomial with leading coefficient #1# is called a monic polynomial.

So a monic quadratic looks like this:

#x^2+bx+c#

If we can find two numbers #alpha# and #beta# such that #alpha+beta = b# and #alphabeta = c#, then we find:

#x^2+bx+c = (x+alpha)(x+beta)#

For example, given:

#x^2+8x+15#

we can find #5+3=8# and #5*3 = 15#, so:

#x^2+8x+15 = (x+5)(x+3)#

What if we find a leading coefficient greater than #1#?
(there's your expression)

This means that we have something like:

#2x^2+7x+6#

How might we try to factor this?

Using an AC method we can try to find a pair of factors of #AC=2*6=12# with sum #B=7# .

The pair #4, 3# works in that #4*3=12# and #4+3=7#.

We can then use this pair to split the middle term and factor by grouping:

#2x^2+7x+6 = (2x^2+4x)+(3x+6)#

#color(white)(2x^2+7x+6) = 2x(x+2)+3(x+2)#

#color(white)(2x^2+7x+6) = (2x+3)(x+2)#

How do you factor when a coefficient is more than 1?