4 Marks Questions

Question 14 Marks

Without actual division, prove that 2x⁴ - 5x³ + 2x² - x + 2 is divisible by x² - 3x + 2.

Answer

Let f(x) = 2x⁴ - 5x³ + 2x² - x + 2
g(x) = x² - 3x + 2
= x² - 2x - x + 2
= x(x - 2) - 1(x - 2)
= (x - 2)(x - 1)
Clearly, (x - 2) and (x - 1) are factors of g(x).
In order to prove that f(x) is exactly divisible by g(x), it is sufficient to prove that f(x) is exactly divisible by (x - 2) and (x - 1).
Thus, we will show that (x - 2) and (x - 1) are factors of f(x).
Now,
f(2) = 2(2)⁴ - 5(2)³ + 2(2)² - 2 + 2
= 32 - 40 + 8 = 0 and
f(1) = 2(1)⁴ - 5(1)³ + 2(1)² - 1 + 2
= 2 - 5 + 2 - 1 + 2 = 0
Therefore, (x - 2) and (x - 1) are factors of f(x).
⇒ g(x) = (x - 2)(x - 1) is a factor of f(x).
Hence, f(x) is exactly divisible by g(x).

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Question 24 Marks

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x⁴ + x³ - 8x² - x + 6, g(x) = 2x - 3

Answer

f(x) = (2x⁴ + x³ - 8x² - x + 6)
By the Factor Theorem, (x - a) will be a factor of f(x) if f(a) = 0.
Here, 2x - 3 = 0
$\text{x}=\frac{3}{2}$
$\text{f}\Big(\frac{3}{2}\Big)=2\Big(\frac{3}{2}\Big)^4+\Big(\frac{3}{2}\Big)^3-8\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)+6$
$=2\times\frac{81}{16}+\frac{27}{8}-8\times\frac{9}{4}-\frac{3}{2}+6$
$=\frac{81}{8}+\frac{27}{8}-18-\frac{3}{2}+6$
$=\frac{81+27-144-12+48}{8}$
$=\frac{156-156}{8}=0$
$\therefore(2\text{x}-3)$ is a factor of $\big(2\text{x}^4+\text{x}^3-8\text{x}^2-\text{x}+6\big).$

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Question 34 Marks

Find the values of a and b so that the polynomial (x³ - 10x² + ax + b) is exactly divisible by (x - 1) as well as (x - 2).

Answer

Let f(x) = (x³ - 10x² + ax + b), then by factor theorem
(x - 1) and (x - 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.
$\therefore$ f(1) = 1³ - 10 × 1² + a × 1 + b = 0
⇒ 1 - 10 + a + b = 0
⇒ a + b = 9 ...(i)
And f(2) = 2³ - 10 × 2² + a × 2 + b = 0
⇒ 8 - 40 + 2a + b = 0
⇒ 2a + b = 32 ...(ii)
Subtracting (i) from (ii), we get
a = 23
Substituting the value of a = 23 in (i), we get
23 + b = 9
⇒ b = 9 - 23
⇒ b = -14
$\therefore$ a = 23 and b = -14

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Question 44 Marks

Without actual division, show that (x³ - 3x² - 13x + 15) is exactly divisible by (x² + 2x - 3).

Answer

Let f(x) = x³ - 3x² - 13x + 15
Now, x² + 2x - 3 = x² + 3x - x - 3
= x (x + 3) - 1 (x + 3)
= (x + 3) (x - 1)
Thus, f(x) will be exactly divisible by x² + 2x - 3 = (x + 3) (x - 1) if (x + 3) and (x - 1) are both factors of f(x), so by factor theorem, we should have f(-3) = 0 and f(1) = 0.
Now, f(-3) = (-3)³ - 3(-3)² - 13(-3) + 15
= -27 - 3 × 9 + 39 + 15
= -27 - 27 + 39 + 15
= -54 + 54 = 0
And, f(1) = 1³ - 3 × 1² - 13 × 1 + 15
= 1 - 3 - 13 + 15
= 16 - 16 = 0
$\therefore$ f(-3) = 0 and f(1) = 0
So, x² + 2x - 3 divides f(x) exactly.

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Question 54 Marks

By actual division, find the quotient and the remainder when (x⁴ + 1) is divided by (x - 1).
Verify that remainder = f(1).

Answer

Let f(x)= x⁴ + 1 and g(x) = x - 1.

Quotient = x³ + x² + x + 1
Remainder = 2
Verification:
Putting x = 1 in f(x), we get
f(1) = 1⁴ + 1 = 1 +1 = 2 = Remainder, when f(x) = x⁴ + 1 is divided by g(x) = x - 1.

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