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Maths 3 Marks Question

Polynomials · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 7 questions.

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3 Marks Question

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Question 13 Marks
Write the cube in expanded form: ${\left( {2x + 1} \right)^3}$
Answer
${\left( {2x + 1} \right)^3}$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$ 
$\therefore {\left( {2x + 1} \right)^3} = {\left( {2x} \right)^3} + {\left( 1 \right)^3} + 3 \times 2x \times 1\left( {2x + 1} \right)$
$ = 8{x^3} + 1 + 6x\left( {2x + 1} \right)\,$
$= 8{x^3} + 12{x^2} + 6x + 1.$
Therefore, the expansion of the expression ${\left( {2x + 1} \right)^3}$ is $8{x^3} + 12{x^2} + 6x + 1$
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Question 23 Marks
Verify that x3 + y3 + z3 - 3xyz = $\frac{1}{2}$(x + y + z)[(x - y)2 + (y- z)2 + (z - x)2].
Answer
L.H.S = x3 + y3 + z3 - 3xyz
= (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
(Using Identity a3+ b3+ c3– 3abc= (a + b + c)(a2+ b2+ c2– ab – bc – ca))
= $\frac{1}{2}$(x + y + z){2(x2 + y2 + z2 - xy - yz - zx)} (Multiplying and Dividing by 2)
= $\frac{1}{2}$(x + y + z)(2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx)
= $\frac{1}{2}$(x + y + z){(x2 - 2xy + y2) + (y2 - 2yz + z2)+ (z2 - 2zx + x2)}
= $\frac{1}{2}$(x + y + z)[(x - y)2 + (y - z)2 + (z - x)2]. (Using Identity (a – b)2= a2– 2ab + b2)
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Question 33 Marks
Use suitable identity to find the product:
$\left( {x + 4} \right)\left( {x + 10} \right)$
Answer
$\left( {x + 4} \right)\left( {x + 10} \right)$
We know that $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$

Here a = 4 and b = 10

We need to apply the above identity to find the product $\left( {x + 4} \right)\left( {x + 10} \right) = {x^2} + \left( {4 + 10} \right)x + \left( {4 \times 10} \right)$
$= {x^2} + 14x + 40.$
Therefore, we conclude that the product$ \left( {x + 4} \right)\left( {x + 10} \right)$ is ${x^2} + 14x + 40$

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Question 43 Marks
Find the remainder when x3 + 3x2 + 3x + 1 is divided by x.
Answer
Let p(x) = x3 + 3x2 + 3x + 1
x.
Remainder   = (0)3 + 3(0)2 + 3(0) + 1 = 1
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Question 53 Marks
Verify $x = - \frac{1}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}$ are zeroes of the polynomial $p\left( x \right) = 3{x^2} - 1$
Answer
$p\left( x \right) = 3{x^2} - 1,\,\,\,\,x = - \frac{1}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}$
We need to check whether $p\left( x \right) = 3{x^2} - 1{\text{ at }}x = - \frac{1}{{\sqrt 3 }},\frac{2}{{\sqrt 3 }}$ is equal to zero or not, i.e., $p\left( {{{ - 1} \over {\sqrt 3 }}} \right)$ and $p\left( {{2 \over {\sqrt 3 }}} \right)$ is equal to zero or not.
At $x = \frac{{ - 1}}{{\sqrt 3 }}$
$p\left( { - \frac{1}{{\sqrt 3 }}} \right) = 3{\left( { - \frac{1}{{\sqrt 3 }}} \right)^2} - 1\,\, = 3\left( {\frac{1}{3}} \right) - 1\,\, = 1 - 1\,\, = 0$
At $x = \frac{2}{{\sqrt 3 }}$
$p\left( {\frac{2}{{\sqrt 3 }}} \right) = 3{\left( {\frac{2}{{\sqrt 3 }}} \right)^2} - 1\,\, = 3\left( {\frac{4}{3}} \right) - 1\, = 4 - 1\, = 3$
Therefore, we can conclude that $x = \frac{{ - 1}}{{\sqrt 3 }}$ is a zero of the polynomial $p\left( x \right) = 3{x^2} - 1$ but $x = \frac{{ - 1}}{{\sqrt 3 }}$ is not a zero of the polynomial $p\left( x \right) = 3{x^2} - 1$
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Question 63 Marks
Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4.
Answer
The zero of x + 2 is –2.
Let p(x) = x3 + 3x2 + 5x + 6 and s(x) = 2x + 4
Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6
= –8 + 12 – 10 + 6
= 0
So, by the Factor Theorem, x + 2 is a factor of x3 + 3x2 + 5x + 6.
Again, s(–2) = 2(–2) + 4 = 0
So, x + 2 is a factor of 2x + 4. 
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Question 73 Marks
Factorise : x3 – 23x2 + 142x – 120
Answer
Let p(x) = x3 – 23x2 + 142x – 120
We shall now look for all the factors of –120. Some of these are $\pm$1, $\pm$2, $\pm$3, $\pm$4, $\pm$5, $\pm$6, $\pm$8, $\pm$10, $\pm$12, $\pm$15, $\pm$20, $\pm$24, $\pm$30, $\pm$60.
By hit and trial, we find that p(1) = 0. Therefore, x – 1 is a factor of p(x).
Now we see that x3 – 23x2 + 142x – 120 = x3 – x2 – 22x2 + 22x + 120x – 120
= x2 (x –1) – 22x(x – 1) + 120(x – 1)
= (x – 1) (x2 – 22x + 120) [Taking (x – 1) common]
Now x2 – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have: x2 – 22x + 120 = x2 – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Therefore, x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)
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