Practice Set 3.4 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials
Question 1
Maharashtra Board Solution
For x = 0, find the value of the polynomial x2 – 5x + 5.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
If p(y) = y2 – 3√2 + 1, then find p( 3√2 ).
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
If p(m) = m3 + 2m2 – m + 10, then P(a) + p(-a) = ?
Solution & Step-by-Step Answer:
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) = a3 – a3 + 2a2 + 2a2 – a + a + 10 + 10 ∴ p(a) + p(-a) = 4a2 + 20
Question 4
Maharashtra Board Solution
If p(y) = 2y3 – 6y2 – 5y + 7, then find p(2).
Solution & Step-by-Step Answer:
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