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Chapter 3 Polynomials Practice Set 3.4

Chapter 3 Polynomials Practice Set 3.4

Question 1.
For x = 0, find the value of the polynomial x² – 5x + 5.
Solution:
p(x) = x² – 5x + 5
Put x = 0 in the given polynomial.
∴ P(0) = (0)² – 5(0) + 5
= 0 – 0 + 5
∴ p(0) = 5

Question 2.
If p(y) = y² – 3√2 + 1, then find p( 3√2 ).
Solution:
p(y) = y² – 3√2 y + 1
Putp= 3√2 in the given polynomial.
∴ p( 3√2 ) = (3√2 )² – 3√2 (3√2 ) + 1
= 9 x 2 – 9 x 2 + 1
= 18 – 18 + 1
∴ p( 3√2 ) = 1

Question 3.
If p(m) = m³ + 2m² – m + 10, then P(a) + p(-a) = ?
Solution:
p(m) = m³ + 2m² – m + 10
Put m = a in the given polynomial.
∴ p(a) = a³ + 2a² – a + 10 …(i)
Put m = -a in the given polynomial.
p(-a) = (-a)³ + 2(-a)² – (-a) +10
∴ p (-a) = -a³ + 2a² + a + 10 …(ii)
Adding (i) and (ii),
p(a) + p(-a) = (a³ + 2a² – a + 10) + (-a³ + 2a² + a + 10)

∴ p(a) + p(-a) = 4a² + 20

Question 4.
If p(y) = 2y³ – 6y² – 5y + 7, then find p(2).
Solution:
p(y) = 2y³ – 6y² – 5y + 7
Put y = 2 in the given polynomial.
∴ p(2) = 2(2)³ – 6(2)² – 5(2) + 7
= 2 x 8 – 6 x 4 – 10 + 7
= 16 – 24 – 10 + 7
∴ P(2) = -11