Polynomials – Expansion & Factorisation Quiz - ICSE Class 10 Mathematics
Free practice quiz: Polynomials – Expansion & Factorisation (ICSE Class 10 Mathematics). 42 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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If $x + \frac{1}{x} = 3$, find the value of $x^3 + \frac{1}{x^3}$.
- 27
- 24
- 21
- 18
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Answer: B. 24
We know that $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. Let $a=x$ and $b=\frac{1}{x}$. Then $(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3x(\frac{1}{x})(x + \frac{1}{x})$. Substituting the given value $x + \frac{1}{x} = 3$, we get $3^3 = x^3 + \frac{1}{x^3} + 3(1)(3)$. So, $27 = x^3 + \frac{1}{x^3} + 9$. Therefore, $x^3 + \frac{1}{x^3} = 27 - 9 = 24$.
Factorize the expression $x^3 - 6x^2 + 11x - 6$.
- $(x-1)(x-2)(x-3)$
- $(x+1)(x-2)(x-3)$
- $(x-1)(x+2)(x-3)$
- $(x-1)(x-2)(x+3)$
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Answer: A. $(x-1)(x-2)(x-3)$
Let $P(x) = x^3 - 6x^2 + 11x - 6$. We can use the Rational Root Theorem to find possible rational roots. Possible rational roots are divisors of the constant term -6, which are $\pm1, \pm2, \pm3, \pm6$. Let's test $x=1$: $P(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. So, $(x-1)$ is a factor. Let's test $x=2$: $P(2) = 2^3 - 6(2)^2 + 11(2) - 6 = 8 - 6(4) + 22 - 6 = 8 - 24 + 22 - 6 = 0$. So, $(x-2)$ is a factor. Let's test $x=3$: $P(3) = 3^3 - 6(3)^2 + 11(3) - 6 = 27 - 6(9) + 33 - 6 = 27 - 54 + 33 - 6 = 0$. So, $(x-3)$ is a factor. Since we have found three roots for a cubic polynomial, the factorization is $(x-1)(x-2)(x-3)$.
Factorize $x^4 - 16y^4$.
- $(x^2-4y^2)(x^2+4y^2)$
- $(x-2y)(x+2y)(x^2+4y^2)$
- $(x^2-2y^2)(x^2+2y^2)$
- $(x-2y)(x+2y)(x^2-4y^2)$
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Answer: B. $(x-2y)(x+2y)(x^2+4y^2)$
The expression $x^4 - 16y^4$ can be written as $(x^2)^2 - (4y^2)^2$. This is a difference of squares, so it factors as $(x^2 - 4y^2)(x^2 + 4y^2)$. The term $x^2 - 4y^2$ can be further factored as $(x)^2 - (2y)^2$, which is also a difference of squares. So, $x^2 - 4y^2 = (x - 2y)(x + 2y)$. Therefore, the complete factorization is $(x - 2y)(x + 2y)(x^2 + 4y^2)$.
If $a+b=7$ and $a^2+b^2=25$, find the value of $ab$.
- 10
- 12
- 14
- 16
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Answer: B. 12
We use the identity $(a+b)^2 = a^2+b^2+2ab$. We are given $a+b=7$ and $a^2+b^2=25$. Substitute these values into the identity: $7^2 = 25 + 2ab$. $49 = 25 + 2ab$. $49 - 25 = 2ab$. $24 = 2ab$. $ab = \frac{24}{2} = 12$.
Factorize $x^3+1$.
- $(x+1)(x^2-x+1)$
- $(x-1)(x^2+x+1)$
- $(x+1)(x^2+x+1)$
- $(x-1)(x^2-x+1)$
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Answer: A. $(x+1)(x^2-x+1)$
This is a sum of cubes expression, $a^3 + b^3 = (a+b)(a^2-ab+b^2)$. Here, $a^3 = x^3 \implies a=x$, and $b^3 = 1^3 \implies b=1$. Substituting into the formula: $(x+1)((x)^2 - (x)(1) + (1)^2)$. $= (x+1)(x^2 - x + 1)$.