Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

## Practice Set 3.4 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

Question 1.

For x = 0, find the value of the polynomial x^{2} – 5x + 5.

Solution:

p(x) = x^{2} – 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) = y^{2} – 3√2 + 1, then find p( 3√2 ).

Solution:

p(y) = y^{2} – 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) = m^{3} + 2m^{2} – m + 10, then P(a) + p(-a) = ?

Solution:

p(m) = m^{3} + 2m^{2} – m + 10

Put m = a in the given polynomial.

∴ p(a) = a^{3} + 2a^{2} – a + 10 …(i)

Put m = -a in the given polynomial.

p(-a) = (-a)^{3} + 2(-a)^{2} – (-a) +10

∴ p (-a) = -a^{3} + 2a^{2} + a + 10 …(ii)

Adding (i) and (ii),

p(a) + p(-a) = (a^{3} + 2a^{2} – a + 10) + (-a^{3} + 2a^{2} + a + 10)

= a^{3} – a^{3} + 2a^{2} + 2a^{2} – a + a + 10 + 10

∴ p(a) + p(-a) = 4a^{2} + 20

Question 4.

If p(y) = 2y^{3} – 6y^{2} – 5y + 7, then find p(2).

Solution:

p(y) = 2y^{3} – 6y^{2} – 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