Arjun Suresh (talk | contribs) |
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==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ==={{Template:Author|Happy Mittal|{{mittalweb}} }}=== | ||
We can use just horner's method, according to which, we can write p(x) as : | We can use just horner's method, according to which, we can write p(x) as : | ||
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| − | $p(x) = a_0 + x(a_1 + x(a_2 + a_3x))$ | + | $$p(x) = a_0 + x(a_1 + x(a_2 + a_3x))$$ |
Consider the polynomial $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ , where $a_i ≠0\; ∀i$. The minimum number of multiplications needed to evaluate p on an input x is:
(A) 3
(B) 4
(C) 6
(D) 9
We can use just horner's method, according to which, we can write p(x) as :
$$p(x) = a_0 + x(a_1 + x(a_2 + a_3x))$$
As we can see, here we need only 3 multiplications, so option (A) is correct.
Note that in question paper, $a_3x^2$ is written instead of $a_3x^3$, but for $a_3x^2$, answer is 2, because we can save one more multiplication between $a_3$ and x, but 2 is not in the options, so I guess question paper had a printing mistake.
Consider the polynomial $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ , where $a_i ≠0\; ∀i$. The minimum number of multiplications needed to evaluate p on an input x is:
(A) 3
(B) 4
(C) 6
(D) 9
We can use just horner's method, according to which, we can write p(x) as :
As we can see, here we need only 3 multiplications, so option (A) is correct.
Note that in question paper, $a_3x^2$ is written instead of $a_3x^3$, but for $a_3x^2$, answer is 2, because we can save one more multiplication between $a_3$ and x, but 2 is not in the options, so I guess question paper had a printing mistake.
GATECSE