Test: Polynomials - 2 - JAMB MCQ

Put x - 1 = 0 
⇒ x = 1
Let F(x) = x⁴ + x³ - 2x² + x + 1
⇒ F(1) = 1⁴ + 1³ - 2 × 1² + 1 + 1
⇒ F(1) = 1 + 1 - 2 + 1 + 1
⇒ F(1) = 4 - 2 = 2
∴Subtracting 2 from  x4 + x³ - 2x² + x + 1 makes it divisible by x - 1.

We have,
f(x) = (k² + 4)x² +13x + 4k
Let α and 1/α are the Zeroes of the polynomial than
Product of zeroes of polynomial = c/a
Here c and a are the constant and coefficient of x² respectively.
⇒ α . 1/α = 4k / k² + 4
⇒ 1 = 4k / k² + 4
⇒ k² + 4 = 4k 
⇒ k² - 4k + 4 = 0
⇒ (k - 2)² = 0
⇒ k = 2, 2
∴ Option 3 is correct.

The factor (x + a) is equal to zero only when its polynomial is also equal to zero.
Even the polynomial may have infinite factors and if any single factor is zero, then the entire polynomial value becomes zero since a value multiplied with a zero is always zero.
Solution
That’s why (x + 4) = 0 and further, that means x = -4.
The correct option is 2.

Degree of the polynomial in 2x⁵ = 5
Degree of the polynomial in 2x³y³ = 6
Degree of the polynomial in 4y⁴ = 4
Degree of the polynomial in 5 = 0
Hence, the highest degree is 6
∴ Degree of polynomial = 6

Degree of the polynomial in 4x⁴ = 4
Degree of the polynomial in 3x³ = 3
Degree of the polynomial in 2x² = 2
Degree of the polynomial in x = 1
Hence, the highest degree is 4.
∴ Degree of polynomial = 4

a = 3 + 2√2
1/a = 1/(3 + 2√2)
⇒ 1/a = (3 – 2√2)/{(3 + 2√2) × (3 – 2√2)}
⇒ 1/a = (3 – 2√2)/{3² – (2√2)²}
⇒ 1/a = (3 – 2√2)/(9 – 8)
⇒ 1/a = (3 – 2√2)
Now,
a + 1/a = 3 + 2√2 + 3 – 2√2
⇒ a + 1/a = 6
(a⁶ – a⁴ – a² + 1)/a³
⇒ a³ – a – 1/a + 1/a³
⇒ (a³ + 1/a³) – (a + 1/a)
⇒ {(a + 1/a)³ – 3(a + 1/a)} – (a + 1/a)
⇒ (6³ – 3 × 6) – 6
⇒ 216 – 18 – 6
⇒ 192
∴ The required value of (a⁶ – a⁴ – a² + 1)/a³ is 192

We know that
x³ + y³ = (x + y)(x² + y² – xy)
Now we have x³ + y³ = 22 and x + y = 5
⇒ 22 = 5(x² + y² – xy)
⇒ 22 = 5[(x + y)² − 3xy)]
⇒ 22 = 5[(5)² − 3xy)]
⇒ xy = 103/15
Now multiply x³ + y³ = 22 with x + y = 5
⇒ x⁴ + y⁴ + xy(x² + y²) = 110
⇒ x⁴ + y⁴ = 110 – xy{(x² + y² − 2xy + 2xy)}
⇒ x⁴ + y⁴ = 110 – xy{(x + y)² − 2xy}
xy = 103/15 and x + y = 5
⇒ x⁴ + y⁴ = 110 – 103/15{(5)² − 2 × 103/15}
⇒ x⁴ + y⁴ = 110 – 6.87{(25 –  13.73}
⇒ x4 + y4 = 110 – 6.87 {(11.27)}
⇒ x⁴ + y⁴ = 110 – 77.42
⇒ x4 + y4 = 32.58
∴ Value of x⁴ + y⁴ is 33.

Let p(x) = 5x³ + 5x² – 6x + 9 
Since, (x + 3) divide p(x), then, remainder will be p(-3).
⇒ p(-3) = 5 × (-3)³ + 5 × (-3)² – 6 × (-3) + 9
⇒ p(-3) = -63

We have, x + 2 = x - (-2)
So, by remainder theorem, when p(x) is divided by (x + 2) = (x - (-2)) the remainder is equal to p(-2).
Now, p(x) = 2x⁵ + 4x⁴ + 7x³ - x² + 3x + 12
⇒ p(-2) = 2(-2)⁵ + 4(-2)⁴ + 7(-2)³ - (-2)² + 3(-2) + 12
⇒ p(-2) = -2(32) + 4(16) - 7(8) - (4) - 6 + 12
⇒ p(-2) = -64 + 64 - 56 - 4 - 6 + 12
⇒ p(-2) =  -66 + 12
⇒ p(-2) = -54
Hence, required remainder = -54.

A quadratic polynomial is a polynomial of degree 2. It has the general form ax² + bx + c, where a, b, and c are constants. Among the given options, only option A, 3x² + 2x + 1, is a quadratic polynomial since it is of degree 2 and has the required form.