SE Class 10 Maths - Factor theorem and its applications

[PaperStSoap]

factor completely

2x³ - 8a²x + 24x² + 72x

so far i got

2x(x² + 12x + 36) - 8a²x
2x(x+6)(x+6) - 8a²x

don't know what to do from here.

 

[Jameson]

factor completely

2x³ - 8a²x + 24x² + 72x

so far i got

2x(x² + 12x + 36) - 8a²x
2x(x+6)(x+6) - 8a²x

don't know what to do from here.

Hi PaperStSoap! Welcome to MHB :)

You are correct the first step is to factor out 2x from every term, but you didn't include the a^2 term in the factoring. It should be:

[math]2x(x^2-4a^2+12x+36)=2x \left(\left[x^2+12x+36 \right]-4a^2 \right)[/math]

Now as you noticed part of this immediately factors: [math]x^2+12x+36=(x+6)(x+6)=(x+6)^2[/math]

So now we have [math]2x \left(\left[x+6 \right]^2-4a^2 \right)[/math]

We can rewrite [math]4a^2[/math] as [math] (2a)^2[/math]

So simplifying once again we have:

[math]2x \left(\left[x+6 \right]^2-(2a)^2 \right)[/math]

This is just a difference of squares though! If we let [math]c=(x+6)[/math] and d = [math]2a[/math] we can think of this as [math]2x(c^2-d^2)=2x(c+d)(c-d)[/math] So how do we get the final answer from here?

[sp][math]2x \left(x+6+2a \right) \left(x+6-2a \right)[/math][/sp]

 

[CaptainBlack]

factor completely

2x³ - 8a²x + 24x² + 72x

so far i got

2x(x² + 12x + 36) - 8a²x
2x(x+6)(x+6) - 8a²x

don't know what to do from here.

Another method:

First rearrange by collecting similar powers of \(x\) and taking out the obvious factoe od \(2x\):

\[2x^3-8a^2x+24x^2+71x=2x^3+24x^2+(71-8a^2)x=2x[x^2+12x+(36-4a^2)]\]

Now the quadratic in the square brackets on the right can be factored by finding its roots using the quadratic formula and constructing the corresponding linear factors.

Spoiler

The roots of \(x^2+12x+(36-4a^2)\) are: \(-6\pm2a\), so:

\[2x^3-8a^2x+24x^2+71x=2x[x^2+12x+(36-4a^2)]=2x(x+6-2a)(x+6+2a)\]

CB