Expansion and factorisation

Question

I have a little problems with a few questions here and I need help.. Thanks ...

1. Factorise completely

$$9x^4 - 4x^2 - 9x^2y^2 + 4y^2 $$

My workings ..

$$ (3x^2+2x)(3x^2-2x) - y^2 (9x^2-4) = (3x^2 + 2x)(3x^2 -2x) - y^2 (3x+2)(3x-2) $$

2. Factorise $3x^2 + 11x - 20$ and , hence Factorise completely

$$11a - 11b - 20 + 3a^2 + 3b^2 - 6ab$$

My workings ...

$$ 11(a-b) - 20 + (3a-3b)(a-b)$$

3. Evaluate the following by algebraic expansion of factorisation

(A) $78^2 + 312 + 4$

(B) $501^2 - 1002 + 1$

Thanks a lot !

Answer 1

HINTS :

For 1.

The coefficients are $\pm 9,\pm 4$, so $$9x^4-4x^2-9x^2y^2+4y^2=(9x^4-9x^2y^2)-(4x^2-4y^2)$$ is worth trying.

For 2.

Compare $3x^2+11x-20$ with $$11a-11b-20+3a^2+3b^2-6ab=11(a-b)-20+3(?)=3(?)+11(a-b)-20$$

For 3.

Note that $312=78\times 4$ and that $1002=2\times 501$.

Answer 2

1. $$9x^4 - 9x^2y^2 - 4x^2 + 4y^2$$ Group in paris such that; $9x^2(x^2-y^2) - 4(x^2 - y^2) = (x^2 - y^2)(9x^2 - 4)$ Then using the difference of 2 squares we get: $$(x-y)(x+y)(3x-2)(3x+2)$$

2.$3x^2 + 11x - 20$, factorise to $(x-5)(3x-4)$. As @mathlove pointed out $11(a-b) - 20 + 3(?)^2$, where $ ? = a-b$. Which is similar to the previous equation. So utilising it's result we get it factorised down to $(a-b-5)(3(a-b-4)$

3. $78^2 + 312 + 4 = 78^2 + 316$, using the fact that $(a + b)^2 = a^2 + 2ab + b^2$, we get $(70 + 8)^2 + 316 = (4900 + 1120 + 64) + 316 = 6400$

4. Apply the same reasoning as number 3