Practice Test: Polynomials - Grade 10 MCQ

Explanation: Sum of zeroes = α + β =√2
Product of zeroes = αβ = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the polynomial is;
x² – (α+β)x + αβ
= x² – (√2)x + (1/3)
= 3x² - 3√2x +1

x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x - 4)(x + 2)
Therefore, x = 4, -2.

For equal roots, discriminant will be equal to zero.
b² - 4ac = 0
b² = 4ac
ac = b²/4
ac > 0 (as square of any number cannot be negative)

If p(x) is a polynomial of degree one, it can be written as:
p(x) = mx + c
where m and c are constants, and m ≠ 0.
Given that p(a) = 0, substituting x = a into p(x):
p(a) = m(a) + c = 0
This equation implies that a is a root or zero of the polynomial p(x), as it satisfies p(a) = 0.
Hence, the correct answer is:
A: Zero of p(x)

The Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n zeros when considering multiplicity and complex numbers. Therefore, the maximum number of distinct zeros it can have is n.

Hence, the correct answer is D.

The quadratic equation x² - 3x - 10 = 0 is the simplest or standard form that has roots -2 and 5. The quadratic equation for a pair of  roots is unique.

• x squared minus 27 = 0 leads to x squared = 27.
• Square root of 27 = square root of (9×3) = 3× square root of 3.
• Hence x = ±3× square root of 3, matching option 2.
• Options ±9× square root of 3 and ±7× square root of 3 would not zero the polynomial.

Given that 2 is the zero of the quadratic polynomial x² + 3x + k.

⇒ (2)² + 3(2) + k = 0

⇒ 4 + 6 + k = 0

⇒ k = -10

A quadratic polynomial in terms of the zeroes α and β is given by

x² - (sum of the zeroes) x + (product of the zeroes)

i.e, f(x) = x² -(α + β) x + αβ

Now,

Given that zeroes of a quadratic polynomial are -3 and 4

Let α = -3 and β = 4

Therefore, substituting the value α = -3 and β = 4 inf(x) = x² -(α + β) x + αβ, we get

f(x) = x² - ( -3 + 4) x +(-3)(4)

= x² - x - 12, which is equal to (x²/2) – (x/2) – 6

Given quadratic polynomial is x² + 99x + 127.

By comparing with the standard form, we get;

a = 1, b = 99 and c = 127

a > 0, b > 0 and c > 0

We know that in any quadratic polynomial, if all the coefficients have the same sign, then the zeroes of that polynomial will be negative.

Therefore, the zeroes of the given quadratic polynomial are negative.