Factorise $9x^2 + 24xy + 16y^2$ completely. Answer ... Hence factorise $144a^8 - (9a^8 + 24a^4 y + 16y^2)$ completely. Answer ...

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Worked solution

1. Identify the perfect square trinomial and factorise it completely.

$9x^2+24xy+16y^2=(3x)^2+2(3x)(4y)+(4y)^2=(3x+4y)^2$

Answer

$(3x+4y)^2$

1. Use part (a) to rewrite the bracket as a perfect square.

From part (a), $9x^2+24xy+16y^2=(3x+4y)^2$ .

Let $x=a^4$ :
$9a^8+24a^4y+16y^2=(3a^4+4y)^2$
So
$144a^8-(9a^8+24a^4y+16y^2)=(12a^4)^2-(3a^4+4y)^2.$

2. Apply difference of two squares and simplify the factors.

Using $A^2-B^2=(A+B)(A-B)$ ,

egin{aligned} (12a^4)^2-(3a^4+4y)^2 &=(12a^4-(3a^4+4y))(12a^4+(3a^4+4y))\\ &=(9a^4-4y)(15a^4+4y). \end{aligned}

Answer

$(9a^4-4y)(15a^4+4y)$

AI analysis — may contain mistakes.

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