Test: Quadratic Polynomials - Grade 12 MCQ

Given:
If 5m² + 22m - 15 = (px + q) (rx + s), p > r,
Calculation:
⇒ 5m² + 22m - 15 = (px + q) (rx + s),
⇒ 5m² + 25m - 3m - 15 = (px + q) (rx + s)
⇒ 5m(m + 5) -3(m + 5) = (px + q) (rx + s)
⇒ (5m - 3) (m + 5) = (px + q) (rx + s)
Here, p = 5, q = -3, r = 1, s = 5
⇒ ps/qr = 5 × 5/-3 × 1 = -25/3
Hence, It is a negative fraction.

Concept used:
(a – b)² = a² – 2ab + b² 
(a + b)² = a² + 2ab + b² 
(a + b) (a – b) = a² – b² 
(-m) x (-n) = +mn
Solution:
(a + b)² = a² + b² = False
(a – b)² = a² – b² = False
(a + b) (a – b) = a² – b² = True
The product of two negative terms is a negative term = False

Concept used:
(a – b)² = a² – 2ab + b² 
(a + b)² = a² + 2ab + b² 
Solution:
(a + b)² + (a – b)² = a² + 2ab + b² + a² – 2ab + b² = 2a² + 2b² 

Hence, the correct option is 3.

Concept:
If α and β are the zeros of polynomial p(x) then,
p(α) = 0 & p(β) = 0  
Calculation:
Let p(x) =  (k - 1)x² + kx +1
According to question, x = -3 is one of its zeros, than
p(x) at x = -3 become zero.
Therefore,
(k - 1)(-3)² + k(-3) +1 = 0
⇒ 9k - 9 - 3k + 1 = 0
⇒ 6k = 8
⇒ k = 4/3
Hence, option 2 is correct.

Given
4x⁴ + 3x³ + 2x² + x + 1
Concept
The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients.
Solution
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

Given:
a = 3 + 2√2
Concept Used:
a² – b² = (a – b)(a + b)
a³ + b³ = (a + b)³ – 3ab(a + b)
Calculation:
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

Concept:
Remainder Theorem: Let p(x) be any polynomial of degree greater than or equal to one and 'a' be any real number. If p(x) is divided by (x - a), then the remainder is equal to p(a).
Calculation:
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.

Concept used:
Remainder theorem: 
If a polynomial p(x) is divided by (x−a), then the remainder is a
constant given by p(a).
Calculation:
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

Formulas used:

(a + b)² = (a² + b² + 2ab)

(a - b)(a + b) = a² - b²

Calculation:

= 25 - (x² + y² + 2xy)

= 5² - (x + y)²

= (5 + x + y) (5 - (x + y))

∴ Factors are ( 5 + x + y ) (5 - x - y)

Concept used:
a² – 2ab + b² = (a – b)²
Solution:
On comparing X²/4 - 2x + 4 with a2 – 2ab + b² 
a = x/2
b = 2
Therefore, its factors = (x/2 - 2) and (x/2 - 2)
Hence, the correct option is 2