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Consider the polynomial $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ , where $a_i \neq 0$, $\forall i$. The minimum number of multiplications needed to evaluate $p$ on an input $x$ is:

1. 3
2. 4
3. 6
4. 9

we can factorize the equation (x+r1)(x+r2)(x+r3), where r1,r2 and r3 are root of equation
so 3 multiplication
Hello reena

Doesn't (x+r1)(x+r2)(x+r3) need 2 multiplication ?

Why're you sure enough that coefficient of $x^{3}$ in our general given equation would be 1 ?

apply the Horner's Rules

P(x)= a0 + a1x + a2x^2 + a3x^3

P(x)= a0 +(( a1+a2x + a3x^2) x )                             // 1 multipication taking the x common

P(x)= a0 +(( a1+(a2 + a3x ) x ) x                          // 2 multipication in x in inner bracket

P(x)= a0 +( ( a1+(a2 + a3x ) x ) x )                  // 3 multipication entire bracket

## Minimum 3 Multipication required .

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 three multiplications, so option (A) is correct.
by

Sir, if the question would be-:

Total number of arithmetic operation required,then answer would be $6(3+3)$.right?

https://gateoverflow.in/2045/gate2014-3-11

@sourav I agree, it should be 6.
a_0+x(a_1+x(a_2+a_3x)) so 3 multiplications required

mul= pair of brackets

p(x)=a0+x(a1+x(a2+a3(x)))

by

3 or 6 ?
3 multiplications