Polynomials - Olympiad Level MCQ, Class 9 Mathematics - Class 9 MCQ

x + ¹/_x = 4 → ( x + ¹/_x)² 
= 16 → x² + ¹/_x² + 2
= 16 → x² +¹/x² = 14
Now,
x² + ¹/_x² = 14 → ( x² + ¹/_x²)²
= 196 → x⁴ + ¹/_x⁴ + 2
= 196 → x⁴ +¹/x⁴ 
= 194.

According to the Factor Theorem, if x + 2 is a factor of the polynomial P(x) = x³ - 2ax² + 16, then substituting x = -2 into P(x) should yield zero.

We can compute P(-2) as follows:

P(-2) = (-2)³ - 2a(-2)² + 16

Now, we simplify each term:

• (-2)³ = -8
• (-2)² = 4, thus the second term becomes -2a × 4 = -8a

Therefore, we can write:

P(-2) = -8 - 8a + 16

Now, we combine like terms:

8 - 8a = 0

To solve for a:

• -8a = -8
• a = 1

Therefore, the value of a is 1.

To solve this, calculate each term separately:

• 3³ = 27
• 5³ = 125
• 8³ = 512

Now sum them as per the expression:

• 27 + 125 - 512 = 152 - 512 = -360

Thus, the correct answer is B.

Given the equations x + y = 5 and xy = 6, we can find x³ + y³ using the identity:

x³ + y³ = (x + y)(x² - xy + y²).

First, we need to calculate x² + y² using the identity:

x² + y² = (x + y)² - 2xy.

Substituting the known values:

• x² + y² = 5² - 2 × 6
• 5² = 25
• 2 × 6 = 12
• x² + y² = 25 - 12 = 13

Now, substitute x² + y² and xy back into the sum of cubes formula:

x³ + y³ = (5)(13 - 6)

• 13 - 6 = 7
• x³ + y³ = (5)(7) = 35

2^x - 2^x-1 = 16 ↔ 2^x - 2^x_/2 = 16 ↔ a - a_/2 = 16 ↔ a_/2 = 16
↔ a =32 ↔ 2^x = 2⁵ ↔ x = 5.
.`. x² = 5² = 25. 

P(x) = x² + 3Qx – 2Q
∵  (x – 2) is a factor of P(x).
∴  P(2) = 0
⇒  (2)² + 3Q × 2 – 2Q = 0
⇒  4 + 6Q – 2Q = 0
⇒  4Q + 4 = 0
⇒  4Q = – 4 ⇒ Q = –1

Given that (x + 2) is a factor, by the Factor Theorem, substituting (x = -2) into the polynomial should yield zero. Compute each term:

• (-2)³ = -8
• -2a * (-2)² = -2a * 4 = -8a

Adding these with the constant term:

• -8 - 8a + 16 = 0

Simplifying this gives:

• 8 - 8a = 0
• This implies a = 1.

By the Factor Theorem, substituting x = a into the polynomial should equal zero.

The polynomial can be represented as:

P(a) = a³ - 3a * a² + 2a² * a + b.

Simplifying each term, we get:

• - a³
• - 3a³
• + 2a³

Combining terms, we find:

P(a) = (a³ - 3a³ + 2a³) + b = 0 + b.

Setting P(a) = 0, we conclude:

b = 0.

Given that (x + 2) and (x - 1) are factors of the polynomial x³ + 10x² + mx + n, their product (x + 2)(x - 1) = x² + x - 2 must also be a factor. To find the remaining linear term, we need to divide the cubic polynomial by this quadratic:

(x³ + 10x² + mx + n) ÷ (x² + x - 2)

Performing polynomial long division:

• Step 1: Divide x³ by x² to get x. Multiply back and subtract to obtain a new dividend of 9x² + (m + 2)x + n.
• Step 2: Divide 9x² by x² to get 9. Multiply back and subtract to get (m - 7)x + (n + 18).

For there to be no remainder:

• Coefficient of x: m - 7 = 0 ⇒ m = 7
• Constant term: n + 18 = 0 ⇒ n = -18

Thus, we find that m = 7 and n = -18.

To find the value of the expression 16x² + 24x + 9 when x = -3/4:

First, substitute x = -3/4 into each term:

• Calculate x²: (-3/4)² = 9/16
• Multiply by 16: 16 × 9/16 = 9
• Calculate the linear term: 24 × -3/4 = -18

Now, add all terms together:

9 + (-18) + 9 = 0

Thus, the value of the expression is 0.

Given the polynomial division:

x³ - 2x² + px - q = (x² - 2x - 3) · Q(x) + (x - 6)

Assume Q(x) is a linear polynomial, say Q(x) = ax + b. Expanding the right-hand side:

(x² - 2x - 3)(ax + b) = ax³ + (b - 2a)x² + (-3a - 2b)x - 3b

Adding the remainder x - 6:

ax³ + (b - 2a)x² + (-3a - 2b + 1)x + (-3b - 6)

Equating coefficients with x³ - 2x² + px - q:

• For x³: a = 1.
• For x²: b - 2(1) = -2 → b = 0.
• For x: -3(1) - 2(0) + 1 = p → p = -2.
• For the constant term: -3(0) - 6 = -q → q = 6.

Thus, p = -2 and q = 6, which corresponds to option C.

After thorough consideration and various methods, we concluded that:

x⁴ - 11x²y² + y⁴ cannot be expressed as a perfect square of any polynomial in x and y with real coefficients.

Difference of Squares Attempt: Applying the difference of squares formula incorrectly:

• (x² - 25)(x² + 25) = x⁴ - 625. This does not equal x⁴ + 625.

Sum of Squares Factoring: Recognising that x⁴ + 625 can be expressed using the identity:

• a⁴ + b⁴ = (a² + ab√2 + b²)(-a² + ab√2 - b²).
• However, this involves irrational coefficients.

Sophie Germain Identity: Applying the identity:

• a⁴ + 4b⁴ = (a² + 2ab + 2b²)(a² - 2ab + 2b²).
• Adjusting for our case where b = 5:
• x⁴ + 625 = (x² + 10x + 25)(x² - 10x + 25).
• This successfully factors into real quadratics with integer coefficients.

To factorise x² - 8x - 20, we need to find two numbers that:

• Multiply to -20
• Add to -8

The two numbers that satisfy these conditions are 2 and -10. Thus, the quadratic factors as:

(x + 2)(x - 10).

To factor the quadratic expression x² - 11xy - 60y², we look for two binomials in the form (x + ay)(x + by) such that:

• The product of the coefficients of y gives -60.
• The sum of these coefficients gives -11.

The numbers 15 and -4 satisfy these conditions because:

• 15 × -4 = -60
• 15 + -4 = -11

Thus, the correct factors are (x - 15y)(x + 4y), which corresponds to option B.

To factorise 216x³ - 64y³, recognise it as a difference of cubes since both terms are perfect cubes.

• Let a = 6x and b = 4y. Then, the expression becomes (6x)³ - (4y)³.
• Apply the difference of cubes formula: a³ - b³ = (a - b)(a² + ab + b²).
• Substitute a and b: (6x - 4y)((6x)² + (6x)(4y) + (4y)²).

Simplify each term:

• 6x - 4y = 2(3x - 2y)
• (6x)² = 36x²
• (6x)(4y) = 24xy
• (4y)² = 16y²

Factor out the common factor from 36x² + 24xy + 16y²: 4(9x² + 6xy + 4y²).

Combine all factors: 8(3x - 2y)(9x² + 6xy + 4y²). This matches option D.

To factorise the expression (x + y)³ − (x − y)³, we can use the identity for the difference of cubes or expand each term separately.

Expanding both terms:

• (x + y)³ = x³ + 3x²y + 3xy² + y³
• (x − y)³ = x³ - 3x²y + 3xy² - y³

Subtracting the second expansion from the first:

(x + y)³ − (x − y)³ = [x³ + 3x²y + 3xy² + y³] − [x³ - 3x²y + 3xy² - y³]

Simplifying:

• x³ - x³ + 3x²y + 3x²y + 3xy² - 3xy² + y³ + y³
• = 6x²y + 2y³

Factorising the simplified expression:

= 2y(3x² + y²)

Thus, the correct answer is option A: 2y(3x² + y²).

To factor the polynomial x³ - 7x + 6, we first find its roots by testing possible integer values.

• Testing x = 1: 1³ - 7(1) + 6 = 0. So, x = 1 is a root, and (x - 1) is a factor.

Next, we divide the polynomial by (x - 1) using synthetic division:

• Coefficients: 1 (for x³), 0 (for x²), -7 (for x), and 6.

Synthetic division setup with root 1 yields a quotient of x² + x - 6.

Factoring x² + x - 6, we find two numbers that:

• Multiply to -6
• Add to 1

These numbers are 3 and -2:

• x² + x - 6 = (x + 3)(x - 2).

Thus, the complete factorization is: (x - 1)(x + 3)(x - 2), which corresponds to option C.

To find the remainder when x¹³ + 1 is divided by x + 1, we can use the Remainder Theorem. The theorem states that the remainder of a polynomial f(x) divided by x - a is f(a). Here, a = -1.

Substituting x = -1 into the polynomial:

f(-1) = (-1)¹³ + 1 = -1 + 1 = 0

Thus, the remainder is 0.

To determine the remainders R₁ and R₂ when the polynomials are divided by x - 2, we can use the Remainder Theorem, which states that the remainder of a polynomial f(x) divided by x - a is f(a).

• For the first polynomial f(x) = x³ + 2x² + 2x - 4:
  • R₁ = f(2) = (2)³ + 2(2)² + 2(2) - 4
  • = 8 + 8 + 4 - 4 = 16
• For the second polynomial g(x) = x³ + 2x² - 3x + 6:
  • R₂ = g(2) = (2)³ + 2(2)² - 3(2) + 6
  • = 8 + 8 - 6 + 6 = 16

Since both remainders R₁ and R₂ are equal to 16, the correct statement is: R₁ = R₂.