2 Marks Questions

Question 12 Marks

Give an example of a binomial of degree 8.

Answer

A polynomial having two term is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x⁸ - 3x.

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Question 22 Marks

Using factor theorem, show that g(x) is a factor of p(x), when

$\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$ $\text{g}(\text{x})=\text{x}+\sqrt2$

Answer

$\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$

By the factor theorem, (x - a) will be factor of p(x) if p(a) = 0.

Here, $\text{f}(-\sqrt{2})=2\sqrt2(-\sqrt2)^2+5(-\sqrt2)+\sqrt2$

$=2\sqrt2\times2-5\sqrt2+\sqrt2$

$=4\sqrt2-5\sqrt2+\sqrt2$

$=5\sqrt2-5\sqrt2=0.$

$\therefore(\text{x}+\sqrt2)$ is a factor of $2\sqrt2\text{x}^2+5\text{x}+\sqrt2.$

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Question 32 Marks

Give an example of a monomial of degree 0.

Answer

A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5.

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Question 42 Marks

Find the value of a for which (x + 1) is a factor of (ax³ + x² - 2x + 4a - 9).

Answer

Let p(x) = ax³ + x² - 2x + 4a - 9
It is given that (x + 1) is a factor of p(x).
⇒ p(-1) = 0
⇒ a(-1)³ + (-1)² - 2(-1) + 4a - 9 = 0
⇒ -a + 1 + 2 + 4a - 9 = 0
⇒ 3a - 6 = 0
⇒ 3a = 6
⇒ a = 2

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Question 52 Marks

Verify that:
1 and 2 are the zeros of the polynomial p(x) = x² - 3x + 2.

Answer

p(x) = x² - 3x + 2 = (x - 1)(x - 2)
⇒ p(1) = (1 - 1) × (1 - 2)
= 0 × (-1)
= 0
Also,
p(2) = (2 - 1)(2 - 2)
= (-1) × 0
= 0
Hence, 1 and 2 are the zeroes of the given polynomial.

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Question 62 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x⁴ + 9x³ + 6x² - 11x - 6, g(x) = x - 1

Answer

f(x) = (2x⁴ + 9x³ + 6x² - 11x - 6)
By the Factor Theorem, (x - 1) will be a factor of f(x) if f(1) = 0.
Here, f(1) = 2 × 1⁴ + 9 × 1³ + 6 × 1² - 11 × 1 - 6
= 2 + 9 + 6 - 11 - 6
= 17 - 17 = 0
$\therefore$ (x - 1) is factor of (2x⁴ + 9x³ + 6x² - 11x - 6).

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Question 72 Marks

Give an example of a trinomial of degree 4.

Answer

A polynomial having three term is called a trinomial. Since the degree of required binomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x⁴ - 3x + 5.

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Question 82 Marks

Find the value of a for which (x - 4) is a factor of (2x³ - 3x² - 18x + a).

Answer

f(x) = (2x³ - 3x² - 18x + a)
x - 4 = 0
⇒ x = 4
$\therefore$ f(4) = 2(4)³ - 3(4)² - 18 × 4 + a
= 128 - 48 - 72 + a
= 128 - 120 + a
= 8 + a
Given that (x - 4) is a factor of f(x).
By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.
⇒ f(4) = 8 + a = 0
⇒ a = -8

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Question 92 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x³ + 7x² - 24x - 45, g(x) = x - 3

Answer

f(x) = (2x³ + 7x² - 24x - 45)
By the Factor Theorem, (x - 3) will be a factor of f(x) if f(3) = 0.
Here, f(3) = 2 × 3³ + 7 × 3² - 24 × 3 - 45
= 54 + 63 - 72 - 45
= 117 - 117 = 0
$\therefore$ (x - 3) is a factor of (2x³ + 7x² - 24x - 45).

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Question 102 Marks

Find the value of a for which (x + 2a) is a factor of (x⁵ - 4a²x³ + 2x + 2a + 3).

Answer

Let p(x) = x⁵ - 4a²x³ + 2x + 2a + 3
It is given that (x + 2a) is a factor of p(x).
⇒ p(-2a) = 0
⇒ (-2a)⁵ - 4a²(-2a)³ + 2(-2a) + 2a + 3 = 0
⇒ -32a⁵ - 4a²(-8a³) - 4a + 2a + 3 = 0
⇒ -32a⁵ + 32a⁵ -2a + 3 = 0
⇒ 2a = 3
$\Rightarrow\text{a}=\frac{3}{2}$

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Question 112 Marks

Find the value of a for which the polynomial (x⁴ - x³ - 11x² - x + a) is divisible by (x + 3).

Answer

Let p(x) = x⁴ - x³ - 11x² - x + a
It is given that p(x) is divisible by (x + 3).
⇒ (x + 3) is a factor of p(x).
⇒ p(-3) = 0
⇒ (-3)⁴ - (-3)³ - 11(-3)² - (-3) + a = 0
⇒ 81 + 27 - 99 + 3 + a = 0
⇒ 12 + a = 0
⇒ a = -12

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Question 122 Marks

Find the value of k for which (x - 1) is a factor of (2x³ + 9x² + x + k).

Answer

f(x) = (2x³ + 9x² + x + k)
x - 1 = 0
⇒ x = 1
$\therefore$ f(1) = 2 × 1³ + 9 × 1² + 1 + k
= 2 + 9 + 1 + k
= 12 + k
Given that (x - 1) is a factor of f(x).
By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.
⇒ f(1) = 12 + k = 0
⇒ k = -12.

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Question 132 Marks

Rewrite the following polynomial in standard form.
x - 2x² + 8 + 5x³

Answer

8 + x - 2x² + 5x³ is a polynomial in standard form as the powers of x are in ascending order.

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Question 142 Marks

Verify that:
0 and 3 are the zeros of the polynomial, r(x) = x² - 3x.

Answer

r(x) = x² - 3x
r(0) = 0² - 3 × 0
Also,
r(3) = 3² - 3 × 3
= 9 - 9
= 0
Hence, 0 and 3 are the zeroes of the given polynomial.

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Question 152 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
$\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6,\ $ $\text{g}(\text{x})=\text{x}-\sqrt2$

Answer

$\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6$
By the factor theorem, (x - a) will be factor of p(x) if(a) = 0.
Here, $\text{f}(\sqrt{2})=7(\sqrt{2})^2-4\sqrt{2}\times\sqrt{2}-6$
$=14-8-6$
$=14-14=0$
$\therefore(\text{x}-\sqrt2)$ is a factor of $\big(7\text{x}^2-4\sqrt2\text{x}-6\big).$

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Question 162 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x³ + 9x² - 11x - 30, g(x) = x + 5

Answer

f(x) = 2x³ + 9x² - 11x - 30
By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.
Here, f(-5) = 2(-5)³ + 9(-5)² - 11(-5) - 30
= -250 + 225 + 55 - 30
= -280 + 280 = 0
$\therefore$ (x + 5) is a factor of (2x³ + 9x² - 11x - 30).

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Question 172 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 69 + 11x - x² + x³, g(x) = x + 3

Answer

By the factor theorem, g(x) = x + 3 will be a factor of p(x) if p(-3) = 0.
Now, p(x) = 69 + 11x - x² + x³
⇒ p(-3) = 69 + 11(-3) - (-3)² + (-3)³
= 69 - 33 - 9 - 27
= 0
Hence, g(x) = x + 3 is a factor of the given polynomial p(x).

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Question 182 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x⁴ - x² - 12, g(x) = x + 2

Answer

f(x) = (x⁴ - x² - 12)
By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.
Here, f(-2) = (-2)⁴ - (-2)² - 12
= 16 - 4 - 12
= 16 - 16 = 0
$\therefore$ (x + 2) is a factor of (x⁴ - x² - 12).

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Question 192 Marks

Verify that:
2 and -3 are the zeros of the polynomial q(x) = x² + x - 6.

Answer

q(x) = x² + x - 6
⇒ q(2) = 2² + 2 - 6
= 4 - 4
= 0
Also,
q(-3) = (-3)² + (-3) - 6
= 9 - 9
= 0
Hence, 2 and -3 are the zeroes of the given polynomial.

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Question 202 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x³ - 8, g(x) = x - 2

Answer

f(x) = (x³ - 8)
By the Factor Theorem, (x - 2) will be a factor of f(x) if f(2) = 0.
Here, f(2) = (2)³ - 8
= 8 - 8 = 0
$\therefore$ (x - 2) is a factor of (x³ - 8).

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Question 212 Marks

Give an example of a monomial of degree 5.

Answer

A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x⁵.

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