3 Marks Question

Question 13 Marks

What must be added to 3x³ + x² - 22x + 9 so that the result is exactly divisible by 3x² + 7x - 6?

Answer

We know that, Dividend = Divisor × Quotient + Remainder
Dividend = 3x³ + x² - 22x + 9
Divisor = 3x² + 7x - 6

Remainder = -2x - 3
So, -(-2x - 3) = 2x + 3 should be added to 3x³ + x² - 22x + 9 to make it exactly divisible by 3x² + 7x - 6.

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Question 23 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{g(x)}=3\text{x}^2-2,\text{x}=\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$

Answer

$\text{g(x)}=3\text{x}^2-2,\text{x}=\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$
We know that
$\text{g(x)}=3\text{x}^2-2$
Given that,
$\text{x}=\Big(\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}\Big)$
Substitute $\text{x}=\frac{2}{\sqrt{3}}$ in g(x)
$\text{g}\Big(\frac{2}{\sqrt{3}}\Big)=3\Big(\frac{2}{\sqrt{3}}\Big)^2-2$
$=3\Big(\frac{4}{3}\Big)-2$
$=4-2$
$=2\neq0$
Now, Substitute $\text{x}=-\frac{2}{\sqrt{3}}$ in g(x)
$\text{g}\Big(\frac{-2}{\sqrt{3}}\Big)=3\Big(\frac{-2}{\sqrt{3}}\Big)^2-2$
$=3\Big(\frac{4}{3}\Big)-2$
$=4-2$
$=2\neq0$
Since, the results when
$\text{x}=\Big(\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}\Big)$ are not 0, they are roots of 3x² - 2

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Question 33 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=5\text{x}-\pi,\text{x}=\frac{4}{5}$

Answer

$\text{f(x)}=5\text{x}-\pi,\text{x}=\frac{4}{5}$
we know that,
$\text{f(x)}=5\text{x}-\pi$
Given that, $\text{x}=\frac{4}{5}$
Substitute the value of x in f(x)
$\text{f}\Big(\frac{4}{5}\Big)=5\Big(\frac{4}{5}\Big)-\pi$
$=4-\pi$
$\neq0$
Since, the result is not equal to zero, $\text{x}=\frac{4}{5}$ is not the root of the polynomial $5\text{x}-\pi$

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Question 43 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=2\text{(x)}+1,\text{x}=\frac{1}{2}$

Answer

$\text{f(x)}=2\text{(x)}+1,\text{x}=\frac{1}{2}$
We know that,
$\text{f(x)}=2\text{(x)}+1$
Given that $\text{x}=\frac{1}{2}$
Substitute the value of x and f(x)
$\text{f}\Big(\frac{1}{2}\Big)=2\Big(\frac{1}{2}\Big)+1$
$=1+1$
$=2\neq0$
Since, the result is not equal to zero
$\text{x}=\frac{1}{2}$ is the root of 2x + 1

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Question 53 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
f(x) = x² - 1, x = 1, -1

Answer

f(x) = x² - 1, x = (1, -1)
we know that,
f(x) = x² - 1
Given that x = (1, -1)
substitute x = 1 in f(x)
f(1) = 1² - 1
= 1 - 1
= 0
Now, substitute x = (-1) in f(x)
f(-1) = (-1)² - 1
= 1 - 1
= 0
Since, the results when x = (1, -1) are 0 they are the roots of the polynomial f(x) = x² - 1

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Question 63 Marks

In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
f(x) = 4x³ - 12x² + 14x - 3, g(x) = 2x - 1

Answer

Here,
f(x) = 4x³ - 12x² + 14x - 3
g(x) = 2x - 1
From, the remainder theorem when f(x) is divided by $\text{g(x)}=\Big(\text{x}-\frac{1}{2}\Big),$ the remainder will be equal to $\text{f}\Big(\frac{1}{2}\Big).$
Let, g(x) = 0
⇒ 2x - 1 = 0
⇒ 2x = 1
$\Rightarrow\ \text{x}=\frac{1}{2}$
Substitute the value of x in f(x)
$\text{f}\Big(\frac{1}{2}\Big)=4\Big(\frac{1}{2}\Big)^3-12\Big(\frac{1}{2}\Big)^2+14\Big(\frac{1}{2}\Big)-3$
$=4\Big(\frac{1}{8}\Big)-12\Big(\frac{1}{4}\Big)+4\Big(\frac{1}{2}\Big)-3$
$=\Big(\frac{1}{2}\Big)-3+7-3$
$=\Big(\frac{1}{2}\Big)+1$
Taking L.C.M.
$=\Big(\frac{2+1}{2}\Big)$
$=\Big(\frac{3}{2}\Big)$
Therefore, the remainder is $\Big(\frac{3}{2}\Big).$

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Question 73 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = 3x³ + x² - 20x +12, g(x) = 3x - 2

Answer

Let g(x) = 0
⇒ 3x - 2 = 0
$\Rightarrow\ \text{x}=\frac{2}{3}$
$\text{f}\Big(\frac{2}{3}\Big)=3\Big(\frac{2}{3}\Big)^3+\Big(\frac{2}{3}\Big)^2-20\Big(\frac{2}{3}\Big)+12$
$=\frac{24}{27}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{24+12-360+324}{27}$
$=\frac{360-360}{27}$
$=\frac{0}{27}$
$=0$
$\because\ \text{f}\Big(\frac{2}{3}\Big)=0,$ by factor theorem, 3x - 2 is a factor of f(x).

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Question 83 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = 2x³ - 9x² + x + 12, g(x) = 3 - 2x

Answer

Let g(x) = 0
⇒ 3 - 2x = 0
⇒ 3 = 2x
$\Rightarrow\ \text{x}=\frac{2}{3}$
$\text{f}\Big(\frac{3}{2}\Big)=2\Big(\frac{3}{2}\Big)^3+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)+12$
$=2\Big(\frac{27}{8}\Big)+9\Big(\frac{9}{4}\Big)-\Big(\frac{3}{2}\Big)+12$
$=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+12$
$=\frac{27-81+6+48}{4}$
$=\frac{81-81}{4}$
$=\frac{0}{4}$
$=0$
$\because\ \text{f}\Big(\frac{3}{2}\Big)=0,$ by factor theorem, 3 - 2x is a factor of f(x).

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Question 93 Marks

If x = 2 is a root of the polynomial f(x) = 2x² - 3x + 7a, Find the value of a.

Answer

We know that, f(x) = 2x² - 3x + 7a
Given that x = 2 is the root of f(x)
Substitute the value of x in f(x)
f(2) = 2(2)²- 3(2) + 7a
= (2 × 4) - 6 + 7a
= 8 - 6 + 7a
= 7a + 2
Now, equal 7a + 2 to zero
⟹ 7a + 2 = 0
⟹ 7a = -2
⟹ a = -27
The value of $\text{a}=-\frac{2}{7}$

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Question 103 Marks

If x - 2 is a factor of the following two polynomials, find the values of a in case:
x⁵ - 3x⁴ - ax³ + 3ax² + 2ax + 4

Answer

x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4
Let,
x - 2 = 0
$\therefore$ x = 2
$\because$ x - 2 is a factor of p(x) = x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4
$\therefore$ p(2) = 0
p(2) = 2⁵ - 3(2)4 - a(2)³ + 3 × a × 2² + 2 × 2 × a + 4 = 0
⇒ 32 - 48 - 8a + 12a + 4a + 4 = 0
⇒ 8a - 12 = 0
⇒ 8a = 12
$\Rightarrow\ \text{a}=\frac{12}{8}=\frac{3}{2}$

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Question 113 Marks

If x - 2 is a factor of the following two polynomials, find the values of a in case:
x³ - 2ax² + ax - 1

Answer

x³ - 2ax² + ax - 1
Let,
x - 2 = 0
$\therefore$ x = 2
$\because$ x - 2 is a factor of p(x) = x³ - 2ax² + ax - 1
$\therefore$ p(2) = 0
p(2) = 2³ - 2a(2)² + 2a - 1 = 0
⇒ 8 - 8a + 2a - 1 = 0
⇒ 7 - 6a = 0
⇒ 6a = 7
$\Rightarrow\ \text{a}=\frac{7}{6}$

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Question 123 Marks

For what value of a is (x - 5) a factor of x³- 3x² + ax - 10?

Answer

Let,
x - 5 = 0
x = 5
$\because$ (x - 5) is a factor of x³ - 3x² + ax - 10
$\therefore$ f(5) = 0
f(5) = 5³ - 3(5)² + a(5) - 10 = 0
⇒ 125 - 3 × 25 + 5a - 10 = 0
⇒ 125 - 85 + 5a = 0
⇒ 40 + 5a = 0
⇒ 5a = -40
$\Rightarrow\ \text{a}=\frac{-40}{5}=-8$

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Question 133 Marks

Find the value of a such that (x - 4) is a factors of 5x³ - 7x² - ax - 28.

Answer

Let g(x) = x - 4, f(x) = 5x³- 7x² - ax - 28
Let g(x) = 0
⇒ x - 4 = 0
⇒ x = 4,
Since (x - 4) is a factor of f(x).
$\therefore$ f(4) = 0
f(4) = 5(4)³ - 7(4)² - a(4) - 28 = 0
⇒ 5(64) - 7(16) - 4a - 28 = 0
⇒ 320 - 112 - 4a - 28 = 0
⇒ 180 - 4a = 0
⇒ 4a = 180
$\Rightarrow\ \text{a}=\frac{180}{4}=45$

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Question 143 Marks

Find the value k if x - 3 is a factor of k²x³ - kx² + 3kx - k.

Answer

Let g(x) = 0
⇒ x - 3 = 0
⇒ x = 3,
Since (x - 3) is a factor of f(x)
$\therefore$ f(3) = 0
f(3) = k²3³ - k3² + 3k(3) - k = 0
⇒ 27k² - 9k + 9k - k = 0
⇒ 27k² - k = 0
⇒ k(27k - 1) = 0
$\therefore$ k = 0, 27k - 1 = 0
27k = 1
$\text{k}=\frac{1}{27}$
Hence k = 0, $\text{k}=\frac{1}{27}$

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Question 153 Marks

Find the remainder when x³ + 3x³ + 3x + 1 is divided by:
$\text{x}+\pi$

Answer

Here,
f(x) = x³ + 3x² + 3x + 1
By remainder theorem
$\text{x}+\pi=0$
$\Rightarrow\ \text{x}=-\pi$
Substitute the value of x in f(x)
$\text{f}(-\pi)=(-\pi)^3+3(-\pi)^2+3(-\pi)+1$
$=(-\pi)^3+3(-\pi)^2+3(-\pi)+1$

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Question 163 Marks

Find the remainder when x³ + 3x³ + 3x + 1 is divided by:

5 + 2x

Answer

Here,
f(x) = x³ + 3x² + 3x + 1
By remainder theorem
5 + 2x = 0
2x = -5
$\text{x}=\frac{-5}{2}$
substitute the value of x in f(x)
$\text{f}\Big(\frac{-5}{2}\Big)=\Big(\frac{-5}{2}\Big)^3+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+3\Big(\frac{25}{4}\Big)+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1$
Taking L.C.M
$=\frac{-125+150-50+8}{8}$
$=\frac{-27}{8}$

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Question 173 Marks

Factorize the following polynomials:
4x³ + 20x² + 33x + 18 given that 2x + 3 is a factor.

Answer

Let f(x) = 4x³ + 20x² + 33x + 18 be the given polynomial.
Therefore, (2x + 3) is a factor of the polynomial f(x).
Now,
f(x) = 2x²(2x + 3) + 7x(2x + 3) + 6(2x + 3)
= (2x + 3)(2x² + 7x + 6)
= (2x + 3)(2x² +4x + 3x + 6)
= (2x + 3)(2x + 3)(x + 2)
Hence (x + 2), (2x + 3) and (2x + 3) are the factors of polynomial f(x).

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