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Factorize the following

$(i)9x^{2}+4y^{2}+16z^{2}+12xy-16yz-24xz$

$(ii)25x^{2}+16y^{2}+4z^{2}-40xy+16yz-20xz$

$(iii)16x^{2}+4y^{2}+9z^{2}-16xy-12yz+24xz$

Answers (1)

$(i)(3x+2y-4z)(3x+2y-4z)$

Solution

Given, $9x^{2}+4y^{2}+16z^{2}+12xy-16yz-24xz$

This can be written as:

$(3x)^{2}+(2y)^{2}+(4z)^{2}+2(3x)(2y)+2(2y)(-4z)+2(3x)(-4z)$ …(i)

We know that,

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

Comparing equation (i) with RHS of above identity, we get:

a = 3x, b = 2y, c = -4z

So, we get:

$(3x)^{2}+(2y)^{2}+(-4z)^{2}+2(3x)(2y)+2(2y)(-4z)+2(3x)(-4z)$

$=(3x+2y-4z)^{2}$

Hence the factorized form is $(3x+2y-4z)(3x+2y-4z)$

$(ii)(-5x+4y+2z)(-5x+4y+2z)$

Solution

Given, $25x^{2}+16y^{2}+4z^{2}-40xy+16yz-20xz$

This can be written as:

$(-5x)^{2}+(4y)^{2}+(2z)^{2}+2(-5x)(4y)+2(4y)(2z)+2(-5x)(2z)$ …(i)

We know that,

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

Comparing equation (i) with RHS of above identity, we get:

a = -5x, b = 4y, c = 2z

So, we get:

$(-5x)^{2}+(4y)^{2}+(2z)^{2}+2(-5x)(4y)+2(4y)(2z)+2(-5x)(2z)$

$=(-5x+4y+2z)^{2}=(-5x+4y+2z)(-5x+4y+2z)$

Hence the factorized form is $(-5x+4y+2z)(-5x+4y+2z)$

$(iii)(4x-2y+3z)(4x-2y+3z)$

Solution

Given, $16x^{2}+4y^{2}+9z^{2}-16xy-12yz+24xz$

This can be written as:

$(4x)^{2}+(-2y)^{2}+(3z)^{2}+2(4x)(-2y)+2(-2y)(3z)+2(4x)(3z)$ …(i)

We know that,

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

Comparing equation (i) with RHS of above identity, we get:

a = 4x, b = -2y, c = 3z

So, we get:

$(4x)^{2}+(-2y)^{2}+(3z)^{2}+2(4x)(-2y)+2(-2y)(3z)+2(4x)(3z)$

$=(4x-2y+3z)^{2}=(4x-2y+3z)(4x-2y+3z)$

Hence the factorized form is $(4x-2y+3z)(4x-2y+3z)$ .

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