Question

3x^2+27y^2+z^2-18xy+6√3yz-2√3zx​

Answer 1

Answer:

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Answer 2

$\bf 3 {x}^{2} + 27 {y}^{2} + {z}^{2} - 18xy + 6 \sqrt{3} yz - 2 \sqrt{3} zx \\ \bf = ( { - \sqrt{3} x + 3 \sqrt{3}y + z) }^{2}$

Given:

• $3 {x}^{2} + 27 {y}^{2} + {z}^{2} - 18xy + 6 \sqrt{3} yz - 2 \sqrt{3} zx$

To find:

• Factorise the polynomial.

Solution:

Identity to be used:

$\bf ( {a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac$

Step 1:

Rewrite the polynomial, so that identity can be applied.

$3 {x}^{2} + 27 {y}^{2} + {z}^{2} - 18xy + 6 \sqrt{3} yz - 2 \sqrt{3} zx = \\( - { \sqrt{3}x) }^{2} + ( {3 \sqrt{3}y) }^{2} + ( {z}^{2} ) + 2( - \sqrt{3} x)(3 \sqrt{3} y) +2(3 \sqrt{3}y)(z) + 2( - \sqrt{3}x)(z)$

Negative sign should be placed before coefficient of x; because in negative terms -18xy and -2√3zx, x is common.

Step 2:

Compare the terms from identity.

It is clearly seen now

$a = - \sqrt{3} x$

$b = 3 \sqrt{3} y$

and

$c = z$

So,

$3 {x}^{2} + 27 {y}^{2} + {z}^{2} - 18xy + 6 \sqrt{3} yz - 2 \sqrt{3} zx \\= ( { - \sqrt{3} x + 3 \sqrt{3}y + z) }^{2}$

Thus,

$\bf 3 {x}^{2} + 27 {y}^{2} + {z}^{2} - 18xy + 6 \sqrt{3} yz - 2 \sqrt{3} zx \\ \bf= ( { - \sqrt{3} x + 3 \sqrt{3}y + z) }^{2}$

Learn more:

1) factorise 4x^2+9y^2+z^2-12xy+6yz-4xz

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2) Expand the following, using suitable identity (√2x +2y -√3z) ^2

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