2. Find p(0),p(1) and p(2) for each of the following polynomials. p(x) = x2-x+1 (ii) p(v) = 2 + y + 2y2 - y3 (iii) p(z)=23 (iv) p(t) = (1 - 1) (t + 1) (v) p(x) = x2 – 3x+2
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i) p(x) = x²-x+1
=> P(0) = 0²-0+1
=> 0-0+1
=> 1
=>P(1) = 1²-1+1
=> 1 - 1 + 1
=> 1
=> P(2) = 2²-2+1
=> 4 - 2 + 1
=> 3
ii) p(y) = 2 + y + 2y² - y³
=>p(0) = 2+0+2(0)²-0³
=>2 + 0 + 0 - 0
=> 2
=>p(1) = 2 + 1 + 2(1)² - 1³
=> 3 + 2(1) - 1
=> 3 + 2 - 1
=> 4
=> p(2) = 2 + 2 + 2(2)² - 2³
=> 4 + 2(4) - 8
=> 4 + 8 - 8
=> 4
iii) p(z)=2³ – this question is incomplete.
iv) p(t) = (1 - 1) (t + 1)
=>p(0) = (1-1)(0+1)
=> (0)(1)
=> 0
=>p(1) = (1-1)(1+1)
=> (0)(2)
=> 0
=>p(2) = (1-1)(2-1)
=> (0)(1)
=> 1
v) p(x) = x²– 3x+2
=> p(0) = 0² - 3(0)+2
=> 0 - 0 + 2
=> 2
=>p(1)= 1² - 3(1) + 2
=> 1 - 3 + 2
=> 0
=>p(2)=2² - 3(2) + 2
=> 4 - 6 + 2
=> 0
HOPE IT HELPS!!!
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