How to factor perfect square trinomials and difference of squares

Expert intro: When an expression looks almost familiar, the trick is to check whether it matches a square trinomial, a difference of squares, or something that simply does not factor nicely over integers. A quick pattern check saves a lot of time.

Detailed walkthrough

a) $9w^2x^2-24wxyz+16y^2z^2$

This is a perfect square trinomial.

• First term: $9w^2x^2=(3wx)^2$
• Last term: $16y^2z^2=(4yz)^2$
• Middle term: $-24wxyz=-2(3wx)(4yz)$

So the factorization is:

$9w^2x^2-24wxyz+16y^2z^2=(3wx-4yz)^2$

b) $625-(y-5)^2$

This is a difference of squares:

$625=25^2$

So,

$625-(y-5)^2=(25-(y-5))(25+(y-5))$

Simplify each factor:

$=(30-y)(y+20)$

c) $144a^4-499b^4$

Write the first term as a square:

$144a^4=(12a^2)^2$

But $499b^4$ is not a perfect square over the integers, so this expression does not factor nicely using standard integer factoring methods.

So the best answer is:

prime over the integers

d) $25x^2+30x+4$

Check for a perfect square trinomial:

• $25x^2=(5x)^2$
• $4=2^2$
• middle term $30x=2(5x)(2)$

Therefore:

$25x^2+30x+4=(5x+2)^2$

💡 Pitfall guide

A common mistake is expanding the middle term too fast and missing the pattern. For part (b), keep the whole quantity $(y-5)$ together until the last step. For part (c), do not force a factorization just because there are two terms; if neither term forms a clean square, it may stay unfactored over integers.

🔄 Real-world variant

If part (b) were written as $a^2-(y-5)^2$, you would still use difference of squares: $(a-(y-5))(a+(y-5))$. If part (d) had a negative middle term like $25x^2-30x+4$, the answer would become $(5x-2)^2$.

🔍 Related terms

perfect square trinomial, difference of squares, factorization by pattern