Question
Factorize: $x^2 - y^2 - 4xz + 4z^2$
✓
Answer
$x^2-y^2-4 x z+4 z^2$
On rearranging the terms $=x^2-4 x z+4 z^2-y^2$
$=(x)^2-2 \times x \times 2 z+(2 z)^2-y^2$
Using the identity $x^2-2 x y+y^2$
$=(x-y)^2=(x-2 z)^2-y^2$
Using the identity $p^2-q^2=(p+q)(p-q)$
$=(x-2 z+y)(x-2 z-y)$
$\therefore x^2-y^2-4 x z+4 z^2$
$=(x-2 z+y)(x-2 z-y)$
On rearranging the terms $=x^2-4 x z+4 z^2-y^2$
$=(x)^2-2 \times x \times 2 z+(2 z)^2-y^2$
Using the identity $x^2-2 x y+y^2$
$=(x-y)^2=(x-2 z)^2-y^2$
Using the identity $p^2-q^2=(p+q)(p-q)$
$=(x-2 z+y)(x-2 z-y)$
$\therefore x^2-y^2-4 x z+4 z^2$
$=(x-2 z+y)(x-2 z-y)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Construct a histogram for the following data:
→| Monthly School fee (in ₹): | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $150-180$ | $180-210$ | $210-240$ |
| No of schools | $5$ | $12$ | $14$ | $18$ | $10$ | $9$ | $4$ |
It being given that $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$ and $\sqrt{10}=3.162,$ find the value of three places of decimals, of the following: $\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$
→In figure $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ$. $OS$ is another ray lying between rays $OP$ and $OR$. Prove that $\angle\text{ROS}=\frac{1}{2}(\angle\text{QOS}-\angle\text{POS}).$
→Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by: $\text{x}+\pi$
→The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $13 m, 14 m$ and $15 m $. The advertisements yield an earning of Rs. $2000 $per $m ^2$ a year. A company hired one of its walls for $6$ months. How much rent did it pay?
→Find the area of a triangle, two sides of which are $8 \ cm$ and $11 \ cm$ and the perimeter is $32 \ cm.$
→If $a+b=10$ and $a b=21$, find the value of $a^3+b^3$
→If $O$ is the centre of the circle, find the value of $x$ in the following figure:
→Find the cost of laying grass in a triangular field of sides $50 m, 65 m$ and $65 m$ at the rate of $Rs.7$ per $m ^2$.
→$AB$ is a line segment and $P$ is its mid$-$point. $D $and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB.$
Show that:
$i. \triangle DAP \cong \triangle EBP$
$ii. AD = BE$
→Show that:
$i. \triangle DAP \cong \triangle EBP$
$ii. AD = BE$