SE Class 10 Maths - Factor theorem and its applications

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Lo and behold, the quadratic factor on the right is a perfect square, so the quadratic formula is not needed. Instead, notice that the quadratic factor is the square of the expression \(x+6\) and the linear factor is the product of \((x+6)\) and \((x+6)\) minus \(4a^2\). Thus the factorization is\[2x(x^2+12x+36-4a^2)=2x(x+6)^2-8a^2x=2x(x+6+2a)(x+6-2a)\]CB
  • #1
PaperStSoap
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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.
 
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  • #2
PaperStSoap said:
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:

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

Now as you noticed part of this immediately factors: \(\displaystyle x^2+12x+36=(x+6)(x+6)=(x+6)^2\)

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

We can rewrite \(\displaystyle 4a^2\) as \(\displaystyle (2a)^2\)

So simplifying once again we have:

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

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

[sp]\(\displaystyle 2x \left(x+6+2a \right) \left(x+6-2a \right)\)[/sp]
 
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  • #3
PaperStSoap said:
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.

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
 

FAQ: SE Class 10 Maths - Factor theorem and its applications

What is a factor polynomial?

A factor polynomial is a polynomial expression that can be written as a product of two or more simpler polynomial expressions. This means that each term of the polynomial can be factored into smaller expressions.

How do you factor a polynomial?

To factor a polynomial, you can use various methods such as the greatest common factor, grouping, or the quadratic formula. It is important to identify the type of polynomial and choose the appropriate method for factoring.

Why is factoring polynomials important?

Factoring polynomials is important because it allows us to simplify complex expressions and solve equations. It also helps in finding the roots or solutions of a polynomial equation, which is useful in many real-world applications.

Can all polynomials be factored?

No, not all polynomials can be factored. Some polynomials, called prime polynomials, cannot be factored into simpler expressions. However, most polynomials can be factored using the appropriate method.

What is the difference between factoring and expanding a polynomial?

Factoring a polynomial involves breaking down a polynomial into simpler expressions, while expanding a polynomial involves simplifying an expression by multiplying out the terms. Factoring and expanding are inverse operations of each other, meaning that factoring can be undone by expanding and vice versa.

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