Question

factorise x²+y²+z²/4+2xy-yz-zx

Answer 1

This is the full answer of your question. You have to also use the property x²+y²+z²/4+2xy-yz-zx = (x+y+z)².

Answer 2

Answer:

Main Aim:-

To factorise $\boxed{\tt{ {x}^{2}+{y}^{2}+ \frac{ {z}^{2} }{4} + 2xy - yz - zx}}$

Note:-

• Always remember all the identities before you factorise any expression.
• Try to recognise the expression's product type then proceed.

Real Content:-

$\sf{{x}^{2} + {y}^{2} + \frac{ {z}^{2} }{4} + 2xy - yz - zx}$

(Expression as given in the question)

$\sf{ = {(x)}^{2} + {(y)}^{2} + {{( - \frac{z}{2} )}}^{2} + 2(x)(y) + 2(y)( - \frac{z}{2} ) + 2( - \frac{z}{2} )(x) }$

(After we have recognised, we have go that the expression is in the form of a²+b²+c²+2ab+2bc+2ca)

Note:-

• x = a
• y = b
• -(z/2) = c

$\sf{= {[x + y + ( - \frac{z}{2} )]}^{2}}$

(As we know, a²+b²+c²+2ab+2bc+2ca = (a+b+c)², we have written in that format)

$\sf{= {[x + y - (\frac{z}{2} )]}^{2} }$

(Changes can be seen in the symbols)

$\sf\green {= (x + y - \frac{z}{2})(x + y - \frac{z}{2}) }$

(Answer/Factorised form of the expression)

TheAssassin's BRAIN BOOSTER:-

SOME MORE USEFUL IDENTITIES:-

• $\tt{{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}}$
• $\tt{{(a-b)}^{2}={a}^{2}-2ab+{b}^{2}}$
• $\tt{({a}^{2}-{b}^{2})=(a+b)(a-b)}$
• $\tt{(x+a)(x+b)={x}^{2}+(a+b)x+ab}$