Question

Factorise :

$\left(i\right)a^2-(b-c{)}^2(ii\left)25\right(2x-y{)}^2-16(x-2y{)}^2(iii\left)16\right(5x+4{)}^2-9(3x-2{)}^2(iv)9x^2-\frac{1}{16}(v\left)25\right(x-2y{)}^2-4$

Answer 1

$\left(i\right)a^2-(b-c{)}^2=(a-(b-c)\left)\right(a+b-c)[a^2-b^2=(a-b\left)\right(a+b\left)\right]=(a-b+c\left)\right(a+b-c\left)\right(ii\left)25\right(2x-y{)}^2-16(x-2y{)}^2=(5(2x-y){)}^2-\left(4\right(x-2y){)}^2=[5(2x-y)-4(x-2y)\left]\right[5(2x-y)+4(x-2y)]=[10x-5y-4x+8y\left]\right[10x-5y+4x-8y][a^2-b^2=(a-b\left)\right(a+b\left)\right]=(6x+3y\left)\right(14x-13y)=3(2x+y\left)\right(14x-13y\left)\right(iii\left)16\right(5x+4{)}^2-9(3x-2{)}^2=(4(5x+y){)}^2-\left(3\right(3x-2){)}^2=\left[4\right(5x+4)-3(3x-2\left)\right]\left[4\right(5x+4)+3(3x-2\left)\right][a^2-b^2=(a-b\left)\right(a+b\left)\right]=(20x+16-9x+6)[20x+16+9x-6]=[11x+22](29x+10)=11(x+2\left)\right(29x+10\left)\right(iv)9x^2-\frac{1}{16}=(3x{)}^2-{\left(\frac{1}{4}\right)}^2=(3x-\frac{1}{4})(3x+\frac{1}{4})\left[\right(a^2-b^2)=(a-b\left)\right(a+b\left)\right]\left(v\right)25(x-2y{)}^2-4=(5(x-2y){)}^2-2^2=\left[5\right(x-2y)-2]\left[5\right(x-2y)+2][a^2-b^2=(a-b\left)\right(a+b\left)\right]=(5x-10y-2)(5x-10y+2)$