I have the answer in my answer book but I don't know how to work it out.
$2a^2 - a - 3$ ----- $(2a - 3)^2$ ------- $4a^2 - 9$
$a^2b^2 - b^4$-------- $ab^2 + b^3$---------- $ab - b^2$
(I used dashes for spaces)
Can someone please explain in really simple terms (I'm new to factorising and quadratics etc).
On
For this part: 2a^2 - a - 3 ----- (2a - 3)^2 ------- 4a^2 - 9
$(2a - 3)^2 = (2a - 3)*(2a - 3) \ \ \text{remember that: } (x - y)^2 = x^2 + 2xy + y^2 = (x - y)*(x - y)\\ 4a^2 - 9 = (2a - 3)*(2a + 3) \ \ \text{remember that: } x^2 - y^2 = (x - y)*(x + y)\\ 2a^2 - a - 3 \ \text{can be factored as: } (2a - 3)*(a+1) \\ \text{polynomial division might help to get this expression}$
So common factor is $(2a - 3)$.
$$2a^2-a-3=2a^2+2a-3a-3=2a(a+1)-3(a+1)=(2a-3)(a+1)$$ $$(2a-3)^2=(2a-3)(2a-3)$$ $$4a^2-9=(2a)^2-(3)^2=(2a+3)(2a-3)$$
Common factor: $2a-3$
$$a^2b^2-b^4=b^2(a^2-b^2)=b^2(a+b)(a-b)$$ $$ab^2+b^3=b^2(a-b)$$ $$ab-b^2=b(a-b)$$
Common factor: $b(a-b)=ab-b^2$
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