Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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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 : | ||
$$p(x) = a_0 + x(a_1 + x(a_2 + a_3x))$$ | $$p(x) = a_0 + x(a_1 + x(a_2 + a_3x))$$ | ||
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| + | As we can see, here we need only 3 multiplications, so option <b>(A)</b> 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 | 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 | ||
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 : $$p(x) = a_0 + x(a_1 + x(a_2 + a_3x))$$ So 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