Polynomials Test - 12

A polynomial of degree 1 is known as linear polynomial.

A polynomial of degree 2 is known as quadratic polynomial.

A polynomial of degree 3 is known as cubic polynomial.

A polynomial of degree 0 is known as constant polynomial.

Number 0 is known as zero polynomial.

Since the degree of the polynomial 2x² + x + 1 is 2, it is a quadratic polynomial.

The correct answer is C.

A polynomial of degree 1 is known as linear polynomial.

A polynomial of degree 2 is known as quadratic polynomial.

A polynomial of degree 3 is known as cubic polynomial.

A polynomial of degree 0 is known as constant polynomial.

Number 0 is known as zero polynomial.

Since the degree of the polynomial x³ is 3, it is a cubic polynomial.

Thus, the information in alternative D is incorrectly matched.

The correct answer is D.

The given expression is (3m + 4n)³.

It is known that (x + y)³ = x³ + y³ + 3xy (x + y).

Here, x = 3m and y = 4n.

∴(3m + 4n)³

= (3m)³+ (4n)³ + 3(3m)(4n)(3m + 4n)

= 27m³ + 64n³ + 36mn(3m + 4n)

= 27m³ + 64n³ + 108m²n + 144mn²

Thus, theexpanded form of the expression (3m + 4n)³ is 27m³ + 64n³ + 108m²n + 144mn².

The correct answer is D.

The given polynomial is 64x³ + y³ − 8z³ + 24xyz.

64x³ + y³ − 8z³ + 24xyz

= (4x)³ + (y)³ + (−2z)³ − 3(4x)(y)(−2z), which is of the form a³ + b³ + c³ − 3abc, where a = 4x, b = y, and c = −2z.

We know that,

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

∴64x³ + y³ − 8z³ + 24xyz

= (4x)³ + (y)³ + (−2z)³ − 3(4x)(y)(−2z)

= (4x + y − 2z)(16x² + y² + 4z² − 4xy + 2yz + 8xz)

Hence, the correct answer is option A.

The number 997 can be written as (1000 − 3).

∴997³ = (1000 − 3)³

We know that (x − y)³ = x³ − y³ − 3xy (x − y).

⇒ (1000 − 3)³ = 1000³ − 3³ − 3 × (1000) × (3)(1000 − 3)

= 1000000000 − 27 − 3 × 1000 × 3 × 997

Thus, the number 997³ can be expressed as 1000000000 − 27 − 3 × 1000 × 3 × 997.

The correct answer is B.

The highest power of the variable in a polynomial is called the degree of the polynomial.

The highest power of variable x in the polynomial 17x⁵ + 4 is 5. Hence, the degree of this polynomial is 5.

Thus, the information in alternative C is correctly matched.

The correct answer is C.

A polynomial of degree 1 is known as linear polynomial.

A polynomial of degree 2 is known as quadratic polynomial.

A polynomial of degree 3 is known as cubic polynomial.

A polynomial of degree 0 is known as constant polynomial.

Number 0 is known as zero polynomial.

Since the degree of the polynomial 3y + 3 is 1, it is not a cubic polynomial. It is a linear polynomial.

The correct answer is D.

The given polynomial is f(x) = x² − 6x + 8.

It is given that f(a) = −1.

At x = a, f(a) = a² − 6a + 8.

∴f(a) = −1

⇒ a² − 6a + 8 = −1

⇒ a² − 6a + 8 + 1 = 0

⇒ a² − 6a + 9 = 0

⇒ a² − 2 ×a × 3+ 3² = 0

⇒ (a − 3)² = 0 [a² − 2ab + b² = (a − b)²]

⇒ a − 3 = 0

⇒ a = 3

Thus, the value of a is 3.

The correct answer is D.

The highest power of the variable in a polynomial is called the degree of the polynomial.

2x³(x⁵ + 3) + 2 = 2x⁸ + 6x³ + 2

Here, the highest power of the variable x in the polynomial 2x³(x⁵ + 3) + 2 or 2x⁸ + 6x³ + 2 is 8. Hence, the degree of this polynomial is 8.

The highest power of the variable t in the polynomial 3t³ + t + 9 is 3. Hence, the degree of this polynomial is 3.

Difference = 8 − 3 = 5

Thus, the difference between the degrees of the two given polynomials is 5.

The correct answer is C.

According to the remainder theorem, when a polynomial p(x) is divided by a linear polynomial (x − a), the remainder obtained is p(a).

Let f(x) = 2x³ − 9x² + 8x + 15 and g (x) = x² − 10x + 50

Now, the zero of (x − 1) is 1.

It is given that the remainder obtained on dividing the polynomial f(x) by (x − 1) is R₁.

∴ R₁ = f(1)

⇒ R₁ = 2(1)³ − 9(1)² + 8(1) + 15 = 2 − 9 + 8 + 15 = 25 − 9 = 16

The zero of (x − 5) is 5.

It is given that the remainder obtained on dividing the polynomial g(x) by (x − 5) is R₂.

∴ R₂ = g(5)

⇒ R₂ = (5)² − 10(5) + 50 = 25 − 50 + 50 = 25

∴ R₁ − R₂ = 16 − 25 = −9

Thus, the value of R₁ − R₂ is −9.

The correct answer is A.

According to the remainder theorem, when a polynomial p(x) is divided by a linear polynomial (x − a), the remainder obtained is p(a).

Let p(x) = 3x⁴ + 3x³ − 5x² + Px + Q

The zero of the linear polynomial x is 0.

It is given that when p(x) is divided by x, the remainder is −30.

∴p(0) = −30

⇒ 3(0)⁴ + 3(0)³ − 5(0)² + P(0) + Q = −30

⇒ Q = −30

∴ p(x) = 3x⁴ + 3x³ − 5x² + Px − 30

The zero of the linear polynomial (x + 1) is −1.

It is also given that when p(x) is divided by (x + 1), the remainder is −45.

∴p(−1) = −45

⇒ 3(−1)⁴ + 3(−1)³ − 5(−1)² + P(−1) − 30 = −45

⇒ 3 − 3 − 5 − P − 30 = −45

⇒ − 35 − P = −45

⇒ P = 45 − 35 = 10

Thus, the value of P is 10.

The correct answer is C.

The polynomial 3x⁴ − (5x³ + 8x² + 2)x + 7 can be written as

3x⁴ − (5x³ + 8x² + 2)x + 7

= 3x⁴ − 5x⁴ − 8x³ − 2x + 7

= − 2x⁴ − 8x³ − 2x + 7

Thus, the coefficient of x⁴ in the given polynomial is −2.

The correct answer is B.

The given polynomial is p(x) = x² + Sx + T .

According to the remainder theorem, when a polynomial p(x) is divided by a linear polynomial (x − a), the remainder obtained is p(a).

It is given that p(x) leaves 2 as the remainder when it is divided by x.

∴ p(0) = 2

⇒ (0)² + S (0) + T = 2

⇒ T = 2

∴ p(x) = x² + Sx + 2

According to the factor theorem, if f(x) is a polynomial of degree n ≥ 1 and g (x) is a linear polynomial, then g(x) is a factor of f(x) if f(x) = 0 at zero of g (x).

The zero of (x + 2) is −2.

Since (x + 2) is a factor of p(x), we must have p(−2) = 0.

p(−2) = 0

⇒ (−2)² + S (−2) + 2 = 0

⇒ 4 − 2S + 2 = 0

⇒ 6 − 2S = 0

⇒ 2S = 6

⇒ S = 3

Thus, the respective values of S and T are 3 and 2.

Hence, the correct answer is option C.

The given polynomial is p(x) = 8x³ − 7x² − 6x + 5.

At x = 1, p(1) = 8(1)³ − 7(1)² − 6(1) + 5 = 8 − 7 − 6 + 5 = 0

∴p(1) = 0

At x = −1, p(−1) = 8(−1)³ − 7(−1)² − 6(−1) + 5 = − 8 − 7 + 6 + 5 = −4

∴p(−1) = −4

Thus, the relation p(−1) = −4 holds for the given polynomial.

The correct answer is D.

According to the remainder theorem, when a polynomial p(x) is divided by a linear polynomial (x − a), the remainder obtained is p(a).

Here, p(x) = 2x³ − 3x² + kx + 4, which is divided by (x − 1).

Now, the zero of the linear polynomial (x − 1) is 1.

∴ Remainder obtained = p(1) = 7

∴p(1) = 2(1)³ − 3(1)² + k(1) + 4 = 7

⇒ 2 − 3 + k + 4 = 7

⇒ 3 + k = 7

⇒ k = 7 − 3 = 4.

Thus, the value of k is 4.

The correct answer is B.

Let p(x) = x³ + x² − x − 1.

Consider the factors of the constant term −1 i.e., ± 1

By trail, we obtain p(−1) = 0

Therefore, by factor theorem, (x + 1) is a factor of p(x).

We can now arrange the polynomial x³ + x² − x − 1 as

x³ + x² − x − 1

= (x³ + x²) − (x + 1)

= x²(x + 1) − (x + 1)

= (x + 1)(x² − 1)

= (x + 1)(x + 1)(x − 1)

= (x + 1)²(x − 1)

The correct answer is C.

The given expression is 4x² + y² + z² − 4xy − 2yz + 4xz.

4x² + y² + z² − 4xy − 2yz + 4xz

= (2x)² + (−y)² + (z)² + 2(2x)(−y) + 2(−y)(z) + 2(z)(2x), which is of the form

a² + b² + c² + 2ab + 2bc + 2ca, where a = 2x, b = −y, c = z.

It is know that (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Therefore, the given expression can be factorised as

4x² + y² + z² − 4xy − 2yz + 4xz

= (2x)² + (−y)² + (z)² + 2(2x)(−y) + 2(−y)(z) + 2(z)(2x)

= {(2x) + (−y) + z}²

= (2x − y + z)²

Thus, (2x − y + z) is a factor of the given expression.

The correct answer is A.