(2a-3b+4c)²= (2a+(-3b)+4c)² = (2a)²+(-3b)²+(4c)²+2(2a)(-3b)+2(-3b)(4c)+2(4c)(2a)
= 4a²+9b²+16c²+(-12ab)+(-24bc)+16ca
= 4a²+9b²+16c²-12ab-24bc+16ca

(a+b+c)²= a²+ b²+ c²+ 2ab + 2bc + 2ca
(x + 2y + 3z)²= (x)²+(2y)²+(3z)²+2(x)(2y)+2(2y)(3z)+2(3z)(x)
= x² + 4y² + 9z² + 4xy + 12yz + 6xz

Replacing ‘b’ by ‘-b ’ in the expansion
(a+b+c)² = a²+ b²+ c²+ 2ab + 2bc + 2ca
(a+(-b)+c)² = a²+(-b)²+ c²+ 2a(-b) + 2(-b)c+ 2ca
= a²+ b²- 2ab - 2bc + 2ca

Let p(x) = 8x⁴- 2x² + 6x-7
Standard form = 8x⁴+ 0x³- 2x²+ 6x - 7
Co-efficients are 8 0 - 2 6 -7
Q(x) = 2x+1,
To find zero of 2x + 1, put 2x + 1 = 0; 2x = -1, x = \(-\frac{1}{2}\)
Synthetic division

Quotient = \(\frac{1}{2}\)[8x³-4x²+6] = 4x³-2x²+3
Quotient 4x³ - 2x² + 3 is compared with the given quotient 4x³-px²-qx+3
Co-efficients of x² is p = -2
Co-efficients of x is q = 0
Remainder is - 10

Let p(x) = 3x³- 4x²- 5
d(x) = (3x+1)
Standard form of p(x) = 3x³- 4x²+ 0x - 5 and d(x) = 3x + 1

To find the zero of 3x+1;
put 3x + 1 = 0 we get 3x = –1
x = \(-\frac{1}{3}\)
\(3x^3-4x^2-5-\left(x+{1\over 3}\right)\left(3x^2-5+{5\over 3}\right)-{50\over 9}\)
\(3x^3-4x^2-5={(3x+1)\over 3}\times3\left(x^2-{5\over3}x+{5\over9}\right)-\left(50\over9\right)\)
\(=(3x+1)\left(x^2-{5\over3}x+{5\over9}\right)-\left(50\over9\right)\)  (since, p(x) = d(x) q(x) + r
Hence the quotient is \(\left(x^2-{5\over3}x+{5\over9}\right)\) and remainder is \(-50\over 9\)

Area of square = side x side
= (3m +2n −4l) \(\times\)(3m +2n −4l)   [substituting a = 3m, b = 2n, c = -4l]
= (3m+2n-4l)²
We know that, (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
[3m+2n+(-4l)]² = (3m)²+(2n)²+(-4l)²+2(3m)(2n) +2(2n)(-4l)+2(-4l)(3m)
= 9m²+4n²+16l²+12mn-16ln-24lm
Therefore, Area of square = [9m²+4n²+16l²+12mn -16ln - 24lm] sq.units.

Compare with ax² + bx + c
Here, a = 2, b = -15, c = −27
product ac = 2\(\times\) –27 = –54, sum b = –15

The required factors are 3 and -18
∴ we split the middle term as –18x and 3x
2x² - 15x -27 = 2x² - 18x + 3x - 27
= 2x (x − 9) + (3x − 9)
= (x - 9) (2x + 3)
Therefore, (x − 9) (2x + 3) and are the factors of 2x² - 15x - 27

Let \(\frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}\) ...(1)

(x²-y²)³+(y²-z²)³+(z²-x²)³=3(x²-y²)(y²-z²)(z²-x²)
(x-y)³+(y-z)³+(z-x)³=3(x-y)(y-z)(z-x)

=(x+y)(y+z)(z+x)