Maths 1 Marks Question

-10 is a non-zero constant. A non-zero constant term is always regarded as having degree 0.

2x² - 7x - 15 = 2x² - 10x + 3x - 15
= 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5) [by splitting middle term]

3x is monomial of degree 1.

Polynomial 2 - x² + x³ is a cubic polynomial, because maximum exponent of x is 3.

Since the highest power of x is 1, the degree of the polynomial 2x - 1 is 1.

4x² + 20x + 25 = (2x)² + 2 × 2x × 5 + (5)²
= (2x + 5)² [Using identity, a² + 2ab + b² = (a + b)²]

Given, polynomial is
h(y) = 2y
For zero of polynomial, put h(y) = 0
2y = 0
Hence, zero of polynomial is 0.

(999)² = (1000 - 1)² = (1000)² - 2 × (1000) × 1 + 1²
= 1000000 - 2000 + 1
= 998001

y³ - 5y is a polynomial in one variable.

Polynomial 1 + x + x² is a quadratic polynomial, because maximum exponent of x is 2.

x² + 9x + 18 = x² + 6x + 3x + 18
= x(x + 6) + 3(x + 6) = (x + 3)(x + 6) [by splitting middle term]

Polynomial 4 - 5y² is a quadratic polynomial, because maximum exponent of y is 2.

y³(1 - y⁴) = y³ - y⁷ Since the highest power of x is 7, the degree of the polynomial is 7.

Polynomial 3x³ is a cubic polynomial, because maximum exponent of x is 3.

Since the highest power of x is 5, the degree of the polynomial x³ - 9x + 3x⁵ is 5.

Polynomial 2 + x is a linear polynomial, because maximum exponent of x is 1.

True

Solution:

Now,

y² + y - 6 = y² + 3y - 2y - 6 [by splitting middle term]

=y(y + 3) - 2(y + 3)

=(y - 2)(y + 3)

Hence, zeroes of y² + y - 6 are 2 and -3.

Let p(x) = (2x - 5)(2x² - 3x + 1)
= 2x(2x² -3x + 1) - 5(2x² - 3x + 1)
= 4x³ - 6x² + 2x - 10x² + 15x - 5
= 4x³ - 16x² + 17x - 5
Hence, the cofficient of x² in p(x) is -16.

x² + x + 1 is a polynomial in one variable.

84 - 2r - 2r²
= -2(r² + r - 42)
= -2(r² + 7r - 6r -42)
= -2[r(r + 7) - 6(r + 7)]
= -2(r - 6)(r + 7) = 2(6 - r)(r + 7)

x² - 2xy + y² + 1 is a polynomial in three variable.

9y² - 66yz + 121z² = (3y)² - 2 × 3y × 11z + (11z)²
= (3y - 11z)² [Using identity, a² - 2ab + b² = (a - b)²]

 Polynomial t² is a quadratic polynomial, because maximum exponent of t is 2.

The coefficient of x² in 3x - 5 is 0.

Let p(x) = (x - 1)(3x - 4)
= 3x² - 7x + 4
= 3x² - 4x - 3x + 4
Hence, the cofficient of x² in p(x) is 3.

Given, polynomial is
p(x) = 10x - 4x² - 3
On putting x = 0, 1 and -2, respectively in Eq. (i),
we get p(0) = 10(0) - 4(0)² - 3 = 0 - 0 - 3 = -3
p(1) = 10(1) - 4(1)² - 3
= 10 - 4 - 3 = 10 - 7 = 3
and p(-2) =10(-2) - 4(-2)² - 3
= -20 - 4 × 4 - 3 = -20 - 16 - 3= -39
Hence, the values of p(0), p(1) and p(-2) are respectively, -3, 3 and -39.

5x² + 3x - 1 is a trinomial of degree 2.

Given, polynomial is
p(x) = x - 4
For zero of polynomial, put p(x) = x - 4 = 0
⇒ x = 4
Hence, zero of polynomial is 4.

Polynomial 3 is a constant polynomial, because the exponent of variable is 0.

Given, polynomial is p(y) = (y + 2)(y - 2)
On putting y = 0, 1 and -2, respectively in Eq. (i),
We get p(0) = (0 + 2)(0 - 2)= -4
p(1) = (1 + 2)(1 - 2)
= 3 × (-1) = -3
and p(-2) = (-2 + 2)(-2 - 2)
= 0(-4) = 0
Hence, the values of p(0), p(1) and p(-2) are respectively, -4, -3 and 0.

x²⁰ - 7is a binomial of degree 20.

Polynomial y³ - y is a cubic polynomial, because maximum exponent of y is 3.

xy + yz + zx is a polynomial in three variable.

6x² + 7x - 3 = 6x² + 9x - 2x - 3
= 3x(2x + 3) -1(2x + 3) = (3x - 1)(2x + 3) [by splitting middle term]

True

Solution:

Now,

t² - 2t = t(t - 2)

Hence, zeroes of t² - 2t are 0 and 2.