Class 9 Maths: Introduction to Linear Polynomials – Extra Questions

Ans: D

Explanation: The degree of a polynomial is the highest power of the variable in the polynomial. In the polynomial 2x – 1, the variable x has a power of 1 (since x is the same as x^1).
Correct Answer: 1

Ans: D

Explanation: A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, and multiplication, and non-negative integer exponents of variables. The given expression $\sqrt{3}x^2 – 2x$ meets these criteria. The exponents of $x$ are 2 and 1, both non-negative integers. The coefficients are $\sqrt{3}$ and $-2$. The expression is formed using subtraction and multiplication.

Correct Answer: $\sqrt{3}x^2 – 2x$

Ans: A

Explanation: The degree of a polynomial is the highest power of the variable in the polynomial. In the polynomial $p(x) = 3x^4 + 6x + 7$, the terms are $3x^4$, $6x$, and $7$. The powers of $x$ are 4, 1, and 0 (since $7 = 7x^0$). The highest power is 4.
Correct Option: A

Ans: A

Solution: A quadratic function has the form ax^2 + bx + c, where a ≠ 0.

Answer: x^2 + 2x + 1

Ans: C

Solution:
$-r^2 – r + 42 = -(r^2 + r – 42) = -(r+7)(r-6)$
Answer: $-(r+7)(r-6)$

Ans: C

Explanation: First, we evaluate $p(2)$.
$p(2) = (2)^3 – 3(2)^2 – 2(2) + 4 = 8 – 12 – 4 + 4 = -4$.
Next, we evaluate $p(-2)$.
$p(-2) = (-2)^3 – 3(-2)^2 – 2(-2) + 4 = -8 – 12 + 4 + 4 = -12$.
Then, we evaluate $p(0)$.
$p(0) = (0)^3 – 3(0)^2 – 2(0) + 4 = 0 – 0 – 0 + 4 = 4$.
Finally, we compute $p(2) + p(-2) – p(0) = -4 + (-12) – 4 = -4 – 12 – 4 = -20$.

Correct Option: C

Ans: D

Explanation: We need to factor the quadratic expression $21x^2 + 11x – 2$. We can do this by finding two numbers that multiply to $(21)(-2) = -42$ and add up to $11$. The numbers $14$ and $-3$ fit the criteria since $14 \times -3 = -42$ and $14 + (-3) = 11$. Then we can rewrite the expression as:
$21x^2 + 14x – 3x – 2$.
Now, factor by grouping:
$7x(3x + 2) – 1(3x + 2)$
$(7x – 1)(3x + 2)$
Alternatively, we can expand each option and see which one gives us the original expression:
A. $(x-2)(7x+1) = 7x^2 + x – 14x – 2 = 7x^2 – 13x – 2$
B. $(7x + 1)(3x – 2) = 21x^2 – 14x + 3x – 2 = 21x^2 – 11x – 2$
C. $(7x – 1)(3x – 2) = 21x^2 – 14x – 3x + 2 = 21x^2 – 17x + 2$
D. $(7x – 1)(3x + 2) = 21x^2 + 14x – 3x – 2 = 21x^2 + 11x – 2$

Correct Option: D

Ans: A

Explanation: The Remainder Theorem states that if a polynomial $p(x)$ is divided by $x – c$, then the remainder is $p(c)$. In this case, $p(x) = x^3 – 6x^2 + 9x + 3$ and $g(x) = x – 1$. We can identify $c$ as $1$. To find the remainder, we need to evaluate $p(1)$.
$p(1) = (1)^3 – 6(1)^2 + 9(1) + 3 = 1 – 6 + 9 + 3 = 7$.
Therefore, the remainder when $p(x)$ is divided by $g(x)$ is 7.

Correct Answer: 7

Ans: B

Solution:
Let the zeros of the polynomial be $x_1$ and $x_2$. The sum of the zeros is $-\frac{7}{2}$. We can find one zero by factoring: $2x^2 + 7x – 4 = (2x – 1)(x + 4)$.
So, the zeros are $x_1 = \frac{1}{2}$ and $x_2 = -4$.
Using the factored form of the polynomial, we can find the zeros. One zero is $\frac{1}{2}$. Let the other zero be $x$. Then $\frac{1}{2} + x = -\frac{7}{2}$, so $x = -\frac{7}{2} – \frac{1}{2} = -\frac{8}{2} = -4$.
The zeros are $\frac{1}{2}$ and $-4$. The product of the zeros is $\frac{-4}{2} = -2$.
Since we know the zeros are $x_1 = \frac{1}{2}$ and $x_2 = -4$, the other zero is $-4$.

Answer: -4

Ans: C

Explanation: Let $f(x)$ be a polynomial of degree 3, and $g(x)$ be a polynomial of degree 4. We can represent the polynomials as:
$f(x) = a_3x^3 + a_2x^2 + a_1x + a_0$ where $a_3 \neq 0$
$g(x) = b_4x^4 + b_3x^3 + b_2x^2 + b_1x + b_0$ where $b_4 \neq 0$
Then, $f(x) – g(x) = (a_3x^3 + a_2x^2 + a_1x + a_0) – (b_4x^4 + b_3x^3 + b_2x^2 + b_1x + b_0) = -b_4x^4 + (a_3-b_3)x^3 + (a_2-b_2)x^2 + (a_1-b_1)x + (a_0-b_0)$.
Since $b_4 \neq 0$, the degree of $f(x) – g(x)$ is 4.

Correct Option: C

Ans: C

One of the zeroes of the polynomial $2x^2 + 7x – 4$ is $\frac{1}{2}$.

Explanation:

Let $p(x) = 2x^2 + 7x – 4$.
We can factorize the polynomial as follows:
$2x^2 + 7x – 4 = 2x^2 + 8x – x – 4$
$= 2x(x + 4) – 1(x + 4)$
$= (2x – 1)(x + 4)$

To find the zeroes, set $p(x) = 0$:
$(2x – 1)(x + 4) = 0$
This gives us two possible solutions:
$2x – 1 = 0 \Rightarrow x = \frac{1}{2}$
$x + 4 = 0 \Rightarrow x = -4$

Therefore, one of the zeroes of the polynomial is $\frac{1}{2}$.

Ans: D

Explanation:
To find the remainder when a polynomial $P(x)$ is divided by $x-a$, we can use the Remainder Theorem, which states that the remainder is $P(a)$.

1. Dividing by $x$: This is equivalent to dividing by $x-0$. So, we evaluate $P(0)$:
$P(0) = (0)^3 + (0)^2 – 9(0) – 9 = -9$.

2. Dividing by $x+1$: This is equivalent to dividing by $x-(-1)$. So, we evaluate $P(-1)$:
$P(-1) = (-1)^3 + (-1)^2 – 9(-1) – 9 = -1 + 1 + 9 – 9 = 0$.

3. Dividing by $x+2$: This is equivalent to dividing by $x-(-2)$. So, we evaluate $P(-2)$:
$P(-2) = (-2)^3 + (-2)^2 – 9(-2) – 9 = -8 + 4 + 18 – 9 = 5$.

The remainders are $-9$, $0$, and $5$.

Correct Option: D

Ans: C

Solution:
To find the remainder when $p(x) = 3x^4 – 6x^2 + 8x – 2$ is divided by $g(x) = x – 2$, we use the Remainder Theorem.
The Remainder Theorem states that the remainder when a polynomial $p(x)$ is divided by $x – a$ is $p(a)$. In this case, $g(x) = x – 2$, so $a = 2$.

Substitute $x = 2$ into $p(x)$:
$p(2) = 3(2)^4 – 6(2)^2 + 8(2) – 2$
$p(2) = 3(16) – 6(4) + 16 – 2$
$p(2) = 48 – 24 + 16 – 2$
$p(2) = 24 + 16 – 2$
$p(2) = 40 – 2$
$p(2) = 38$

Answer: 38

Ans: C

Solution:
$x^3 + 2x^2 – 9x – 18 = x^2(x+2) – 9(x+2) = (x^2-9)(x+2) = (x-3)(x+3)(x+2)$
Answer: $(x-3)(x+3)(x+2)$

Ans: C

Solution: The polynomial contains two variables, $a$ and $b$.
Answer: Two variables

Ans: D

Solution:
The degree of the zero polynomial is undefined. Some texts define it as -∞.
Answer: Undefined

Ans: B

Explanation: To find possible expressions for the length and width, we need to factor the given area expression, $(14x^2 – 11x – 15)$. We look for two binomials that multiply to give this quadratic. We can try factoring by grouping or by trial and error.

We’re looking for factors of (14 * -15) = -210 that add up to -11. The factors -21 and 10 fit this criteria since -21 + 10 = -11. Thus, we rewrite the middle term and factor by grouping:
$14x^2 – 11x – 15 = 14x^2 – 21x + 10x – 15$
$= 7x(2x – 3) + 5(2x – 3)$
$= (7x + 5)(2x – 3)$.

Now let’s examine the options:
A. $(3x – 2)m$ and $(5x + 8)m$. $(3x-2)(5x+8) = 15x^2 + 24x – 10x – 16 = 15x^2 + 14x -16$. This doesn’t match, so option A is incorrect.
B. $(7x + 5)m$ and $(2x – 3)m$. $(7x+5)(2x-3) = 14x^2 – 21x + 10x – 15 = 14x^2 – 11x – 15$. This matches the original expression.
C. Both A and B. A is incorrect
D. None of these.

Therefore, the correct expression is the factorization found above. Comparing to our options, we note that the factors for B work correctly.

Correct Option: B

Ans: C

An expression in two variables is symmetric if interchanging the variables leaves the expression unchanged. In this case, we have $ax^3 + 2x^2y – bxy^2 – 2y^3$. Swapping $x$ and $y$ gives $ay^3 + 2y^2x – byx^2 – 2x^3$. For the expression to be symmetric, the coefficients of the corresponding terms must be equal. Therefore, comparing the coefficients, we have:
$a = -2$ (from $ax^3$ and $-2x^3$) and $2 = -b$ (from $2x^2y$ and $-bxy^2$), which implies $b=-2$. Thus, $(a,b) = (-2, -2)$. Since the solution given in html does not seem to accurately address how to identify a symmetric polynomial, the response below provides a proper explanation and derivation of the solution.
$(a, b) = (-2, -2)$

Ans: D

Solution:
HCF is the product of the lowest powers of the common variables.
The common variables are x, y, and z.
The lowest power of x is x^2.
The lowest power of y is y.
The lowest power of z is z.
The greatest common factor of the coefficients 15, 20, and 25 is 5.
HCF = $5x^2yz$

Answer: $5x^2yz$

Ans: A

Explanation: Let $u = 3x + y$. Then the expression becomes $5u^2 + 6u – 8$. We can factor this quadratic by looking for two numbers that multiply to $5 \times -8 = -40$ and add to $6$. The two numbers are $10$ and $-4$.
We can rewrite the quadratic as $5u^2 + 10u – 4u – 8$. Factoring by grouping, we have
$5u(u + 2) – 4(u + 2)$
$(5u – 4)(u + 2)$
Now substitute $3x + y$ back in for $u$.
$(5(3x + y) – 4)((3x + y) + 2)$
$(15x + 5y – 4)(3x + y + 2)$

Correct Answer: $(15x + 5y – 4)(3x + y + 2)$