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Any integer can be written in the form 5m , 5m+1, 5m+2 (where m is any integer)
(5m)² = 25m² = 5 × 5m² = 5q (where q = 5m²)
(5m+1)² = (5m)² + 2 ×5m × 1 + 1² [by using identity- (a+b)²= (a²+ 2ab +b² )]
= 25m² + 10 m + 1
= 5 (5m²+ 2m) +1
= 5q +1 ( where q= 5m²+2m)
(5m+2)²= (5m)² + 2× 5m × 2 +2² [by using identity- (a+b)²= (a²+ 2ab +b² )]
= 25m² + 20 m +4
= 5 (5m²+4m ) + 4
= 5q+4 (where q= 5m²+4m)
Hence proved
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Sia ? 6 years, 4 months ago
If g(x) = x2 + 2x + k is a factor of f(x) = 2x4 + x3 - 14x2 + 5x + 6, then remainder is zero when f(x) is divided by g(x).
Let quotient = Q and remainder = R
Let us now divide f(x) by g(x).
R = x(7k + 21) + (2k2 + 8k + 6) -------(1) and Q = 2x2 - 3x - 2(k + 4).------------(2)
Now, R = 0.
{tex}\Rightarrow{/tex} x (7k + 21) + 2 (k2 + 4k + 3) = 0
{tex}\Rightarrow{/tex} 7x (k + 3) + 2 (k+1)(k+3) = 0
{tex}\Rightarrow{/tex} (k+3) [7x + 2(k+1)] = 0
{tex}\Rightarrow{/tex} k + 3 = 0
{tex}\Rightarrow{/tex} k = -3
Thus, polynomial f(x) can be written as,
2x4 + x3 - 14x2 + 5x + 6 = (x2 + 2x + k) [2x2 - 3x - 2(k + 4)] = (x2 + 2x - 3) (2x2 - 3x - 2)
Zeros of x2 + 2x - 3 are,
x2 + 2x - 3 = 0
{tex}\Rightarrow{/tex} (x + 3) (x - 1) = 0
{tex}\Rightarrow{/tex} x = -3 or x = 1
Zeros of (2x2 - 3x - 2) are,
2x2 - 3x - 2 = 0
{tex}\Rightarrow{/tex} 2x2 - 4x + x - 2 = 0
{tex}\Rightarrow{/tex} 2x(x - 2) + 1(x - 2) = 0
{tex}\Rightarrow{/tex} (x - 2)(2x + 1) = 0
x = 2 or x = -{tex}\frac12{/tex}
Thus, the zeros of f(x) are: -3 ,1, 2 and -{tex}\frac12{/tex}
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Nisba Tabassum 7 years, 3 months ago
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