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Maths 4 Marks Questions

Polynomials · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 4 questions.

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Question 14 Marks
Multiply x2 + 4y2 + z2 + 2xy + xz - 2yz by(-z + x - 2y).
Answer
(x2 + 4y2 + z2 + 2xy + xz - 2yz)(-z + x - 2y)

= x2(-z + x - 2y) + 4y2(-z + x - 2y) + z2(-z + x - 2y) + 2xy(-z + x - 2y) + xz(-z + x - 2y) - 2yz(-z + x - 2y)

= x2z + x3 - 2x2y - 4y2z + 4xy2 - 8y3 - z3 + xz2 - 2yz2 - 2xyz + 2x2y - 4xy2 - xz2 + x2z - 2xyz + 2yz2 - 2xyz + 4y2z

= (-x2z + x2z ) + x3 + (-2x2y + 2x2y) + (-4y2z + 4y2z) + (4xy2 - 4xy2)- 8y3 - z3 + (xz2 - xz2) + (-2yz2 + 2yz2) + (-2xyz - 2xyz - 2xyz)

= x3 - 8y3 - z3 - 6xyz

Alternate Answer:

Now, (x - 2y - z)(x2 + 4y2 + z2 + 2xy + xz - 2yz)

= (x - 2y - z)[(x)2 + (-2y)2 + (-z)2 - (x)(-2y) - (-2y)(-z) - (x)(-z)]

= (x3) + (-2y)3 + (-z)3 - 3(x)(-2y)(-z)

[Using identity, a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)]

= x3 - 8y3 - z3 - 6xyz

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Question 24 Marks
Without actual division, prove that 2x4 - 5x3 + 2x2 - x + 2 is divisible by x2 - 3x + 2.
[Hint: Factorise x2 - 3x + 2]
Answer
We have,
x2 - 3x + 2 = x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
Let f(x) = 2x4 - 5x3 + 2x2 - x + 2
Now, p(1) = 2(1)4 - 5(1)3 + 2(1)2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0
p(1) = 0
Therefore, (x - 1) divides p(x)
And p(2) = 2(2)4 - 5(2)3 + 2(2)2 - 2 + 2
= 32 - 40 + 8 - 2 + 2 = 0
p(2) = 0
Therefore, (x - 2) divides p(x).
So, (x - 1)(x - 2) = x2 - 3x + 2 divides 2x2 - 5x3 + 2x2 - x + 2
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Question 34 Marks
The polynomial p(x) = x4 - 2x3 + 3x2 - ax + 3a - 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.
Answer
We know that if p(x) is divided by x + a, then the remainder = p(-a).
Now, p(x) = x4 - 2x3 + 3x2 - ax + 3a - 7 is divided by x + 1, then the remainder = p(-1)
Now, p(-1) = (-1)4 - 2(-1)3 + 3(-1)2 - a(-1) + 3a - 7
= 1 - 2(-1) + 3(1) + a + 3a - 7
= 1 + 2 + 3 + 4a -7
= -1 + 4a
Also, remainder = 19
$\therefore$ -1 + 4a = 19
⇒ 4a = 20; a = 20 ÷ 4 = 5
Again, when p(x) is divided by x + 2, then
Remainder = p(-2) = (-2)4 - 2(-2)3 + 3(-2)2 - a(-2) + 3a -7
= 16 + 16 + 12 + 2a + 3a -7
= 37 + 5a
= 37 + 5(5) = 37 + 25 = 62
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Question 44 Marks
Prove that (a + b + c)3 - a3 - b3 - c3 = 3(a + b )(b + c)(c + a).
Answer
(a + b + c)3 = [a + (b + c)]3

= a3 + 3a2(b + c) + 3a(b + c)2 + (b + c)3

= a3 + 3a2b + 3a2c + 3a(b2 + 2bc + c2) + (b3 + 3b2c + 3bc2 + c3)

= a3 + 3a2b + 3a2c + 3ab2 + 6abc + 3ac2 + b3 + 3b2c + 3bc2 + c3

= a3 + b3 + c3 + 3a2b + 3a2c + 3b2c + 3c2a + 3c2b + 6abc

= a3 + b3 + c3 + 3a2(b + c) + a3 + b3 + c3 + 3a2(b + c)

Hence, above result can be put in the form

(a + b + c)3 = (a + b + c)3 + 3(a + b)(b + c)(c + a)

$\therefore$ (a + b + c)3 - a3 - b3 - c3 = 3(a + b)(b + c)(c + a)

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4 Marks Questions - Maths STD 9 Questions - Vidyadip