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Summation of Polynomials Using Horizontal Method

Simplify:i x-...

Question

Simplify:
(i) (x − y)(x + y) (x² + y²)(x⁴ + y²)
(ii) (2x − 1)(2x + 1)(4x² + 1)(16x⁴ + 1)
(iii) (4m − 8n)² + (7m + 8n)²
(iv) (2.5p − 1.5q)² − (1.5p − 2.5q)²
(v) (m² − n²m)² + 2m³n²

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Solution

To simplify, we will proceed as follows:
(i)
$(i) (x - y) (x + y) (x^{2} + y^{2}) (x^{4} + y^{4}) = (x^{2} - y^{2}) (x^{2} + y^{2}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (x^{4} - y^{4}) (x^{4} + y^{4}) [∵ (a + b) (a - b) = a^{2} - b^{2}] = x^{8} - x^{8} [∵ (a + b) (a - b) = a^{2} - b^{2}]$

$(ii) (2 x - 1) (2 x + 1) (4 x^{2} + 1) (16 x^{4} + 1) = ({(2 x)}^{2} - 1^{2}) (4 x^{2} + 1) (16 x^{4} + 1) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (4 x^{2} - 1) (4 x^{2} + 1) (16 x^{4} + 1) = \{{(4 x^{2})}^{2} - {(1^{2})}^{2}\} (16 x^{4} + 1) [∵ (a + b) (a - b) = a^{2} - b^{2}] = (16 x^{4} - 1) (16 x^{4} + 1) = {(16 x^{4})}^{2} - 1^{2} [∵ (a + b) (a - b) = a^{2} - b^{2}] = 256 x^{8} - 1$

$(iii) {(7 m - 8 n)}^{2} + {(7 m + 8 n)}^{2} = 2 {(7 m)}^{2} + 2 {(8 n)}^{2} [∵ {(a - b)}^{2} + {(a + b)}^{2} = 2 a^{2} + 2 b^{2}] = 98 m^{2} + 128 n^{2}$

$(iv) {(2.5 p - 1.5 q)}^{2} - {(1.5 p - 2.5 q)}^{2} = {(2.5 p)}^{2} + {(1.5 q)}^{2} - 2 (2.5 p) (1.5 q) - [{(1.5 p)}^{2} + {(2.5 q)}^{2} - 2 (1.5 p) (2.5 q)] = {(2.5 p)}^{2} + {(1.5 q)}^{2} - 2 (2.5 p) (1.5 q) - {(1.5 p)}^{2} - {(2.5 q)}^{2} + 2 (1.5 p) (2.5 q) = {(2.5 p)}^{2} - {(1.5 p)}^{2} + {(1.5 q)}^{2} - {(2.5 q)}^{2} = [(2.5 p + 1.5 p) (2.5 p - 1.5 p)] + [(1.5 q + 2.5) (1.5 q - 2.5 q)] [∵ (a + b) (a - b) = a^{2} - b^{2}] = 4 p \times p + 4 q \times (- q) = 4 p^{2} - 4 q^{2} = 4 (p^{2} - 4 q^{2})$

$v) {(m^{2} - n^{2} m)}^{2} + 2 m^{3} n^{2} = {(m^{2})}^{2} + {(n^{2} m)}^{2} [∵ {(a - b)}^{2} + 2 a b = a^{2} + b^{2}] = m^{4} + n^{4} m^{2}$

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