📝 Polynomial functions degree leading coefficient (12 MCQs)

📖 From Calculus • 1. Basics before calculus • 12 questions available

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What is Polynomial functions degree leading coefficient?

Definition:
A polynomial function is a sum of terms $a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ , where the degree is the highest exponent $n$ (with $a_n \neq 0$ ), and the leading coefficient is $a_n$ , which determines end behavior.

Example:
For $f(x) = 3x^4 - 2x^2 + 1$ , degree is 4 (even) and leading coefficient is 3 (positive), so as $x \to \pm\infty$ , $f(x) \to +\infty$ .

Reason:
Degree and leading coefficient predict graph shape and asymptotic behavior, crucial for curve sketching and solving equations.

Easy

Medium

Hard

📝 All Polynomial functions degree leading coefficient MCQs

Q1. What is the degree of the polynomial $3x^4 - 5x^2 + 7$ ?

A.2

B.4 ✅

C.3

D.5

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The degree of a polynomial is the highest exponent of the variable with a non‑zero coefficient. In the expression $3x^4 - 5x^2 + 7$ the term with the largest exponent is $3x^4$ , whose exponent is 4. Hence the polynomial’s degree is 4, which matches option B.

Q2. If a polynomial $p(x)$ has zeros at $x=1$ and $x=-2$ , which of the following must be a factor of $p(x)$ ?

A.(x+1)(x-2)

B.(x-1)^2

C.(x+2)^2

D.(x-1)(x+2) ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Because a zero $r$ of a polynomial implies $(x-r)$ is a factor, the zeros $1$ and $-2$ give the linear factors $(x-1)$ and $(x+2)$ . Their product $(x-1)(x+2)$ must therefore divide $p(x)$ . This corresponds to option D.

Q3. Given $p(x)=x^3-4x^2+5x-2$ , what is $p(2)$ ?

A.2

B.-2

C.0 ✅

D.4

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Evaluating the polynomial at $x=2$ gives $p(2)=2^3-4\cdot2^2+5\cdot2-2=8-16+10-2=0$ . Since the result is zero, the remainder upon division by $x-2$ is zero, and option C correctly states this value.

Q4. Which statement correctly describes the end behavior of the polynomial $f(x)= -2x^5 + 3x^3 - x$ compared to $g(x)= x^5 - 4x^2 + 1$ ?

A.f(x) falls while g(x) rises as

x\to\infty

✅

B.Both rise to

+\infty

x\to\infty

C.Both fall to

-\infty

x\to\infty

D.f(x) rises while g(x) falls as

x\to\infty

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The leading term dictates end behavior. For $f(x)$ the dominant term is $-2x^5$ , which tends to $-\infty$ as $x\to\infty$ . For $g(x)$ the dominant term is $x^5$ , which tends to $+\infty$ . Thus $f(x)$ falls and $g(x)$ rises, matching option A.

Q5. If the product $h(x)=p(x)q(x)$ has a double root at $x=3$ and $p(x)$ is known to have a simple root at $x=3$ , what must be true about $q(x)$ at $x=3$ ?

q(3)=0

and

q&#039;(3)=0

q(3)=0

and

q&#039;(3)\neq0

✅

q(3)\neq0

q(3)=0

only

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A double root means the total multiplicity at $x=3$ is two. Since $p(x)$ contributes a simple (multiplicity 1) root, $q(x)$ must contribute the remaining multiplicity 1. A simple root requires the function to vanish while its derivative does not, giving $q(3)=0$ and $q'(3)\neq0$ , which is option B.

Q6. When $p(x)=x^3-6x^2+11x-6$ is divided by $x-1$ , what is the remainder?

A.1

B.2

C.3

D.0 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The Remainder Theorem states that the remainder of dividing $p(x)$ by $x-a$ equals $p(a)$ . Substituting $a=1$ gives $p(1)=1-6+11-6=0$ . Hence the remainder is 0, which appears as option D.

Q7. Which polynomial grows faster as $x\to\infty$ : $f(x)=3x^4-2x^2$ or $g(x)=5x^3+7x$ ?

A.g(x) grows faster

B.f(x) grows faster ✅

C.Both grow at the same rate

D.Growth cannot be determined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Growth rate is governed by the highest degree term. $f(x)$ has degree 4 while $g(x)$ has degree 3, so as $x$ becomes large, the $x^4$ term dominates and $f(x)$ outpaces $g(x)$ . Therefore option B is correct.

Q8. Which of the following polynomials has zeros at $x= -1, 2$ and a leading coefficient of $4$ ?

4(x-1)(x-2)

4(x+1)(x+2)

4(x+1)(x-2)

✅

4(x-1)(x+2)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A polynomial with specified zeros can be written as the product of linear factors $(x - r_i)$ . For zeros $-1$ and $2$ the factors are $(x+1)$ and $(x-2)$ . Multiplying and then applying the leading coefficient $4$ yields $4(x+1)(x-2)$ , which is option C.

Q9. If a polynomial $p(x)$ is an odd function, which of the following must be true about its coefficients?

A.All odd-degree coefficients are zero

B.Constant term is non‑zero

C.Leading coefficient must be negative

D.All even-degree coefficients are zero ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: An odd function satisfies $p(-x) = -p(x)$ . This condition forces every term with an even power of $x$ to vanish, because such terms would remain unchanged under the sign change. Consequently all even-degree coefficients must be zero, making option D correct.

Q10. Let $q(x)=x^4+ax^3+bx^2+cx+d$ be divided by $x-1$ with remainder $5$ . Which of the following equations must hold?

1+a+b+c+d=5

✅

1-a+b-c+d=5

1+a-b+c-d=5

a+b+c+d=5

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: By the Remainder Theorem, the remainder equals $q(1)$ . Substituting $x=1$ gives $q(1)=1+a+b+c+d$ . Setting this equal to the given remainder 5 yields the equation $1+a+b+c+d=5$ , which matches option A.

Q11. For a monic cubic polynomial $p(x)=x^3+px^2+qx+r$ , which expression gives the sum of its three roots?

-q

-r

-p

✅

p

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Vieta’s formulas relate coefficients to sums of roots. For a monic cubic $x^3+px^2+qx+r$ , the sum of the roots equals $-p$ (the negative of the coefficient of $x^2$ ). Thus the correct expression is $-p$ , which is option C.

Q12. If a polynomial $f(x)$ has a root at $x=2$ of multiplicity 2, what can be said about $f'(2)$ and $f''(2)$ ?

f&#039;(2)=0

and

f&#039;&#039;(2)=0

f&#039;(2)=0

but

f&#039;&#039;(2)\neq0

✅

f&#039;(2)\neq0

and

f&#039;&#039;(2)=0

f&#039;(2)\neq0

and

f&#039;&#039;(2)\neq0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A root of multiplicity m forces the function and its first $m-1$ derivatives to vanish at that point. With multiplicity 2, both the function and its first derivative are zero at $x=2$ , while the second derivative is generally non‑zero. Hence $f'(2)=0$ and $f''(2)\neq0$ , which corresponds to option B.

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