Question

Factorise :

$\left(i\right)15(5x-4{)}^2-10(5x-4\left)\right(ii)3a^{2}x-bx+3a^2-b(iii\left)b\right(c-d{)}^2+a(d-c)+3(c-d)\left(iv\right)a{x}^2+b^{2}y-a{b}^2-x^{2}y\left(v\right)1-3x-3y-4(x+y{)}^2$

Answer 1

$\left(i\right)15(5x-4{)}^2-10(5x-4)=5(5x-4)\left[3\right(5x-4{)}^2-2]=5(5x-4)\left[3\right(25x^2-40x+16)-2]=5(5x-4\left)\right(75x^2-120x+46\left)\right(ii)3a^{2}x-bx+3a^2-b=x(3a^2-b)+1(3a^2-b)=(x+1\left)\right(3a^2-b\left)\right(iii\left)b\right(c-d{)}^2+a(d-c)+3(c-d)=b(c-d{)}^2-a(c-d)+3(c-d)=(c-d)\left[b\right(c-d)-a+3]=(c-d\left)\right(bc-bd-a+3\left)\right(iv)a{x}^2+b^{2}y-a{b}^2-x^{2}y=a{x}^2-a{b}^2+b^{2}y-x^{2}y=a(x^2-b^2)+y(b^2-x^2)=a(x^2-b^2)-y(x^2-b^2)=(x^2-b^2\left)\right(a-y)=(x-b\left)\right(x+b\left)\right(a-y\left)\right(v)1-3x-3y-4(x+y{)}^2=1-3(x+y)-4(x+y{)}^2=1-4(x+y)+(x+y)-4(x+y{)}^2=1[1-4(x+y\left)\right]+(x+y)[1-4(x+y\left)\right]=[1-4x-4y](1+x+y)$