How do you deal with a leading coefficient greater than 1 ?

Jun 2, 2017

A few thoughts...

Explanation:

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:

${x}^{3} + 5 {x}^{2} - 7$

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 ${x}^{3} + 5 {x}^{2} - 7$, the "leading coefficient" is $1$, because we could have equivalently written $1 {x}^{3}$.

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

So a monic quadratic looks like this:

${x}^{2} + b x + c$

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

${x}^{2} + b x + c = \left(x + \alpha\right) \left(x + \beta\right)$

For example, given:

${x}^{2} + 8 x + 15$

we can find $5 + 3 = 8$ and $5 \cdot 3 = 15$, so:

${x}^{2} + 8 x + 15 = \left(x + 5\right) \left(x + 3\right)$

What if we find a leading coefficient greater than $1$?

This means that we have something like:

$2 {x}^{2} + 7 x + 6$

How might we try to factor this?

Using an AC method we can try to find a pair of factors of $A C = 2 \cdot 6 = 12$ with sum $B = 7$.

The pair $4 , 3$ works in that $4 \cdot 3 = 12$ and $4 + 3 = 7$.

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

$2 {x}^{2} + 7 x + 6 = \left(2 {x}^{2} + 4 x\right) + \left(3 x + 6\right)$

$\textcolor{w h i t e}{2 {x}^{2} + 7 x + 6} = 2 x \left(x + 2\right) + 3 \left(x + 2\right)$

$\textcolor{w h i t e}{2 {x}^{2} + 7 x + 6} = \left(2 x + 3\right) \left(x + 2\right)$