Question

Factorise z²-x²-2x-1

Answer 1

Answer:

$\large\boxed{\sf{(z+x+1)(z-x-1)}}$

Step-by-step explanation:

${z}^{2} - {x}^{2} - 2x - 1 \\ \\ = {z}^{2} - ( {x}^{2} + 2x + 1) \\ \\ = {z}^{2} - \left[{(x)}^{2} + 2( x)(1) + {(1)}^{2} \right]\\ \\ = {z}^{2} - {(x + 1)}^{2} \\ \\ = ( z + x + 1)(z - x - 1)$

Concept Map:-

• ${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

• ${x}^{2} - {y}^{2} = (x + y)(x - y)$

Answer 2

{z}^{2} - {x}^{2} - 2x - 1 \\ \\ = {z}^{2} - ( {x}^{2} + 2x + 1) \\ \\ = {z}^{2} - \left[{(x)}^{2} + 2( x)(1) + {(1)}^{2} \right]\\ \\ = {z}^{2} - {(x + 1)}^{2} \\ \\ = ( z + x + 1)(z - x - 1)\end{gathered}z2−x2−2x−1=z2−(x2+2x+1)=z2−[(x)2+2(x)(1)+(1)2]=z2−(x+1)2=(z+x+1)(z−x−1)