Question

Find the zeroes of the polynomial f(x)= px2+(p+q)x+q

Answer 1

Step-by-step explanation:

px^2+px+qx+q=0

px(x+1)+q(x+1)=0

(px+q)(x+1)=0

px+q=0

px=-q

x=-q/p

or

x+1=0

x=-1

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Answer 2

Given:

☛ p(x) = px² + (p + q)x + q

To Find:

☛ Zeroes of the polynomial p(x)

Solution:

☛ p(x) = px² + (p + q)x + q

Zeroes are the constants that satisfies a polynomial and evaluates it to 0.

So,

➜ p(x) = 0

➜ px² + (p + q)x + q = 0

➜ px² + px + qx + q = 0

➜ px( x + 1 ) + q(x + 1) = 0

➜ (x + 1) ( px + q ) = 0

Zeroes:

☛ x + 1 = 0

➜ x = -1

☛ px + q = 0

➜ px = -q

➜ x = -q/p

Hence, Zeroes are -1 and -q/p