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If the following polynomials, when p(x) and q(x) are divided by (x-4), leave the samereminder in each case. Find the value of'k':*p (x)kxcube + 3x2 - 3 & q(x) = 2x3– 5x + k |
Answer» Solution :-→ p(x) = kx³ + 3x² - 3 → p(4) = k(4)³ + 3(4)² - 3 → p(4) = (64k) + 3*16 - 3 → p(4) = (64k) + 48 - 3 → p(4) = (64k) + 45 Similarly, → Q(x) = 2x³ - 5x + k → q(4) = 2(4)³ - 5*4 + k → q(4) = 2*64 - 20 + k → q(4) = 128 - 20 + k → q(4) = 108 + k Now, since remainder is same in each case, → 64k + 45 = 108 + k → 64k - k = 108 - 45 → 63k = 63 → k = 1 (Ans.) Hence, VALUE of k will be 1 . |
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