Factorisation - Class 8 Mathematics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Examples - Factorisation | NCERT | Mathematics | Class 8
Factorise $12 a^{2} b+15 a b^{2}$
To factorize the expression $12a^{2}b + 15ab^{2}$, follow these steps:
Find the Greatest Common Factor (GCF):
Identify the GCF of the coefficients (numeric factors), which in this case is 3.
Also, find the GCF of the variables, which is $a$ (with the lowest exponent of 1) and $b$ (also with the lowest exponent of 1).
Rewrite the Expression Using the GCF:
Factor out the GCF from each term.
Putting it all together:
Step 1: The GCF of $12, 15$ is $3$, and the least powers of $a$ and $b$ present in both terms are $a^1$ and $b^1$, respectively. So, the GCF is $3ab$.
Step 2: Divide each term by $3ab$:
$12a^{2}b$ becomes $4a$, because $12a^{2}b / 3ab = 4a$.
$15ab^{2}$ becomes $5b$, because $15ab^{2} / 3ab = 5b$.
Final Factored Form:$$3ab(4a + 5b)$$
Factorise $10 x^{2}-18 x^{3}+14 x^{4}$
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Factorise $6 x y-4 y+6-9 x$
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Factorise $x^{2}+8 x+16$
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Factorise $4 y^{2}-12 y+9$
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Factorise $49 p^{2}-36$
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Factorise $a^{2}-2 a b+b^{2}-c^{2}$
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Factorise $m^{4}-256$
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Factorise $x^{2}+5 x+6$
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Find the factors of $y^{2}-7 y+12$
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Obtain the factors of $z^{2}-4 z-12$
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Find the factors of $3 m^{2}+9 m+6$
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Do the following divisions.
(i) $-20 x^{4} \div 10 x^{2}$
(ii) $7 x^{2} y^{2} z^{2} \div 14 x y z$
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Divide $44\left(x^{4}-5 x^{3}-24 x^{2}\right)$ by $11 x(x-8)$
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Find the least number of square tiles of equal dimensions required to pave the floor of a room $4.03 \mathrm{m}$ long and $2.17 \mathrm{m}$ wide.
A. 31
B. 91
C. 45
D. 62
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Ask Chatterbot AIExercise 12.1 - Factorisation | NCERT | Mathematics | Class 8
Find the common factors of the given terms
(i) $12 x, 36$
(ii) $2 y, 22 x y$
(iii) $14 p q, 28 p^{2} q^{2}$
(iv) $2 x, 3 x^{2}, 4$
(v) $6 a b c, 24 a b^{2}, 12 a^{2} b$
(vi) $16 x^{3},-4 x^{2}, 32 x$
(vii) $10 p q, 20 q r, 30 r p$
(viii) $3 x^{2} y^{3}, 10 x^{3} y^{2}, 6 x^{2} y^{2} z$
To find the common factors of the given terms, let's examine each item step by step.
(i) Terms: $12x, 36$
Prime factorization $12 = 2^2 \cdot 3$, and $36 = 2^2 \cdot 3^2$.
Common factors: $2^2$ and $3$.
(ii) Terms: $2y, 22xy$
Prime factorization $2 = 2$, and $22 = 2 \cdot 11$.
Common factors: $2$ and incorporating variable, $2y$ if considering y in both.
(iii) Terms: $14pq, 28p^{2}q^{2}$
Prime factorization $14 = 2 \cdot 7$, and $28 = 2^2 \cdot 7$.
Common factors: $2$, $7$, and incorporating variables, $2pq$.
(iv) Terms: $2x, 3x^{2}, 4$
Prime factorization $2 = 2$, $3 = 3$, and $4 = 2^2$.
Common factors: The constant factor is $1$, and for variables, $x$ is common.
(v) Terms: $6abc, 24ab^{2}, 12a^{2}b$
Prime factorization $6 = 2 \cdot 3$, $24 = 2^3 \cdot 3$, and $12 = 2^2 \cdot 3$.
Common factors: $2$ and $3$, for variables $a$ and $b$ are common, so $6ab$.
(vi) Terms: $16x^{3}, -4x^{2}, 32x$
Prime factorization $16 = 2^4$, $4 = 2^2$, and $32 = 2^5$.
Common factors: Considering the sign, just the magnitude, $2^2$ is common, and for variables, $x$. Thus, $4x$.
(vii) Terms: $10pq, 20qr, 30rp$
Prime factorization $10 = 2 \cdot 5$, $20 = 2^2 \cdot 5$, and $30 = 2 \cdot 3 \cdot 5$.
Common factors: No constant number is common across all three, and there's no single variable present in all; hence, no common factors.
(viii) Terms: $3x^{2}y^{3}, 10x^{3}y^{2}, 6x^{2}y^{2}z$
Prime factorization $3 = 3$, $10 = 2 \cdot 5$, and $6 = 2 \cdot 3$.
Common factors: For numbers, no common factor across all three. For variables, $x^2$ and $y^2$ are common.
These steps bring us to the conclusion about the common factors for each set of terms.
Factorise the following expressions
(i) $7 x-42$
(ii) $6 p-12 q$
(iii) $7 a^{2}+14 a$
(iv) $-16 z+20 z^{3}$
(v) $20 l^{2} m+30 a l m$
(vi) $5 x^{2} y-15 x y^{2}$
(vii) $10 a^{2}-15 b^{2}+20 c^{2}$
(viii) $-4 a^{2}+4 a b-4 c a$
(ix) $x^{2} y z+x y^{2} z+x y z^{2}$
(x) $a x^{2} y+b x y^{2}+c x y z$
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Factorise
(i) $x^{2}+x y+8 x+8 y$
(ii) $15 x y-6 x+5 y-2$
(iii) $a x+b x-a y-b y$
(iv) $15 p q+15+9 q+25 p$
(v) $z-7+7 x y-x y z$
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Ask Chatterbot AIExercise 12.2 - Factorisation | NCERT | Mathematics | Class 8
Factorise the following expressions
(i) $a^{2}+8 a+16$
(ii) $p^{2}-10 p+25$
(iii) $25 m^{2}+30 m+9$
(iv) $49 y^{2}+84 y z+36 z^{2}$
(v) $4 x^{2}-8 x+4$
(vi) $121 b^{2}-88 b c+16 c^{2}$
(vii) $(l+m)^{2}-4 l m$
(Hint: Expand $(l+m)^{2}$ first)
(viii) $a^{4}+2 a^{2} b^{2}+b^{4}$
Let's go through the factorization of each expression step by step.
(i) $a^2 + 8a + 16$
This is a perfect square trinomial. The pattern is:
$$ (a+b)^2 = a^2 + 2ab + b^2 $$
Step 1: Identify $a$ and $b$.
Step 2: Apply the pattern.
$$ a^2 + 8a + 16 = (a + 4)^2 $$
(ii) $p^2 - 10p + 25$
This is another perfect square trinomial.
$$ (a - b)^2 = a^2 - 2ab + b^2 $$
$$ p^2 - 10p + 25 = (p - 5)^2 $$
(iii) $25m^2 + 30m + 9$
This is also a perfect square trinomial.
$$ (5m)^2 + 2(5m)(3) + 3^2 = (5m + 3)^2 $$
(iv) $49y^2 + 84yz + 36z^2$
This can be factored as a perfect square trinomial.
$$ (7y)^2 + 2(7y)(6z) + (6z)^2 = (7y + 6z)^2 $$
(v) $4x^2 - 8x + 4$
Another perfect square trinomial.
$$ (2x)^2 - 2(2x)(2) + 2^2 = (2x - 2)^2 $$
(vi) $121b^2 - 88bc + 16c^2$
This also forms a perfect square trinomial.
$$ (11b)^2 - 2(11b)(4c) + (4c)^2 = (11b - 4c)^2 $$
(vii) $(l + m)^2 - 4lm$
This leverages the difference of two squares:
$$ a^2 - b^2 = (a - b)(a + b) $$
Step 1: Expand $(l + m)^2$.
Step 2: Notice $(l + m)^2 = l^2 + 2lm + m^2$.
$$ (l + m)^2 - 4lm = l^2 + 2lm + m^2 - 4lm = l^2 - 2lm + m^2 = (l - m)^2 $$
(viii) $a^4 + 2a^2b^2 + b^4$
This can be identified as a perfect square as well:
$$ (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4 $$
Thus, each expression is factored by recognizing them as perfect squares or by applying algebraic identities.
Factorise
(i) $4 p^{2}-9 q^{2}$
(ii) $63 a^{2}-112 b^{2}$
(iii) $49 x^{2}-36$
(iv) $16 x^{5}-144 x^{3}$
(v) $(l+m)^{2}-(l-m)^{2}$
(vi) $9 x^{2} y^{2}-16$
(vii) $\left(x^{2}-2 x y+y^{2}\right)-z^{2}$
(viii) $25 a^{2}-4 b^{2}+28 b c-49 c^{2}$
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Factorise the expressions
(i) $a x^{2}+b x$
(ii) $7 p^{2}+21 q^{2}$
(iii) $2 x^{3}+2 x y^{2}+2 x z^{2}$
(iv) $a m^{2}+b m^{2}+b n^{2}+a n^{2}$
(v) $(l m+l)+m+1$
(vi) $y(y+z)+9(y+z)$
(vii) $5 y^{2}-20 y-8 z+2 y z$
(viii) $10 a b+4 a+5 b+2$
(ix) $6 x y-4 y+6-9 x$
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Factorise
(i) $a^{4}-b^{4}$
(iv) $x^{4}-(x-z)^{4}$
(ii) $p^{4}-81$
(v) $a^{4}-2 a^{2} b^{2}+b^{4}$
(iii) $x^{4}-(y+z)^{4}$
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Factorise the following expressions
(i) $p^{2}+6 p+8$
(ii) $q^{2}-10 q+21$
(iii) $p^{2}+6 p-16$
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Ask Chatterbot AIExercise 12.3 - Factorisation | NCERT | Mathematics | Class 8
Carry out the following divisions
(i) $28 x^{4} \div 56 x$
(ii) $-36 y^{3} \div 9 y^{2}$
(iii) $66 p q^{2} r^{3} \div 11 q r^{2}$
(iv) $34 x^{3} y^{3} z^{3} \div 51 x y^{2} z^{3}$
(v) $12 a^{8} b^{8} \div\left(-6 a^{6} b^{4}\right)$
To carry out these divisions with steps, we divide both the coefficients and subtract the exponents for like bases:
(i) $28 x^{4} \div 56 x$
Divide the coefficients: $28 \div 56 = 1/2$
Subtract the exponents of like bases: $x^{4} \div x^{1} = x^{4-1} = x^{3}$
So, the result is $\frac{1}{2} x^{3}$.
(ii) $-36 y^{3} \div 9 y^{2}$
Divide the coefficients: $-36 \div 9 = -4$
Subtract the exponents of like bases: $y^{3} \div y^{2} = y^{3-2} = y^{1} = y$
So, the result is $-4y$.
(iii) $66 p q^{2} r^{3} \div 11 q r^{2}$
Divide the coefficients: $66 \div 11 = 6$
For $p$, since there is no matching base in the divisor, it remains as $p$
Subtract the exponents of $q$: $q^{2} \div q = q^{2-1} = q^{1} = q$
Subtract the exponents of $r$: $r^{3} \div r^{2} = r^{3-2} = r^{1} = r$
So, the result is $6 p q r$.
(iv) $34 x^{3} y^{3} z^{3} \div 51 x y^{2} z^{3}$
Divide the coefficients: $34 \div 51 = 2/3$
Subtract the exponents of $x$: $x^{3} \div x = x^{3-1} = x^{2}$
Subtract the exponents of $y$: $y^{3} \div y^{2} = y^{3-2} = y^{1} = y$
For $z$, since the exponents are equal ($z^{3} \div z^{3}$), they cancel out to $1$
So, the result is $\frac{2}{3} x^{2} y$.
(v) $12 a^{8} b^{8} \div\left(-6 a^{6} b^{4}\right)$
Divide the coefficients: $12 \div -6 = -2$
Subtract the exponents of $a$: $a^{8} \div a^{6} = a^{8-6} = a^{2}$
Subtract the exponents of $b$: $b^{8} \div b^{4} = b^{8-4} = b^{4}$
So, the result is $-2 a^{2} b^{4}$.
Divide the given polynomial by the given monomial
(i) $\left(5 x^{2}-6 x\right) \div 3 x$
(ii) $\left(3 y^{8}-4 y^{6}+5 y^{4}\right) \div y^{4}$
(iii) $8\left(x^{3} y^{2} z^{2}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3}\right) \div 4 x^{2} y^{2} z^{2}$
(iv) $\left(x^{3}+2 x^{2}+3 x\right) \div 2 x$
(v) $\left(p^{3} q^{6}-p^{6} q^{3}\right) \div p^{3} q^{3}$
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Work out the following divisions
(i) $(10 x-25) \div 5$
(ii) $(10 x-25) \div(2 x-5)$
(iii) $10 y(6 y+21) \div 5(2 y+7)$
(iv) $9 x^{2} y^{2}(3 z-24) \div 27 x y(z-8)$
(v) $96 a b c(3 a-12)(5 b-30) \div 144(a-4)(b-6)$
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Divide as directed
(i) $5(2 x+1)(3 x+5) \div(2 x+1)$
(ii) $26 x y(x+5)(y-4) \div 13 x(y-4)$
(iii) $52 p q r(p+q)(q+r)(r+p) \div 104 p q(q+r)(r+p)$
(iv) $20(y+4)\left(y^{2}+5 y+3\right) \div 5(y+4)$
(v) $x(x+1)(x+2)(x+3) \div x(x+1)$
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Factorise the expressions and divide them as directed
(i) $\left(y^{2}+7 y+10\right) \div(y+5)$
(ii) $\left(m^{2}-14 m-32\right) \div(m+2)$
(iii) $\left(5 p^{2}-25 p+20\right) \div(p-1)$
(v) $5 p q\left(p^{2}-q^{2}\right) \div 2 p(p+q)$
(iv) $4 y z\left(z^{2}+6 z-16\right) \div 2 y(z+8)$
(vi) $12 x y\left(9 x^{2}-16 y^{2}\right) \div 4 x y(3 x+4 y)$
(vii) $39 y^{3}\left(50 y^{2}-98\right) \div 26 y^{2}(5 y+7)$
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Ask Chatterbot AIExtra Questions - Factorisation | NCERT | Mathematics | Class 8
Evaluate the following expression using patterns:
$$ 35 \times 9999 $$
A) 369965
B) 349935
C) 34935
D) 349965
The correct option to evaluate the expression $$35 \times 9999$$ is D) 349965.
To solve, consider the expression: $$ 35 \times 9999 $$ We can simplify this by rewriting $9999$ as $10000 - 1$ which gives: $$ 35 \times (10000 - 1) $$ Applying the distributive property: $$ 35 \times 10000 - 35 \times 1 $$ Which simplifies to: $$ 350000 - 35 $$ Resulting in: $$ 349965 $$ Thus, the final answer is 349965.
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If $(x+1)$ is a factor of $x^{4}-(p-3)x^{3}-(3p-5)x^{2}+(2p-7)x+6$, then $p=$
A) 4
B) 2
C) 1
D) None of these
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If $x-a$ is a factor of $x^{3}-3x^{2}+2a^{2}x+b$, then the value of $b$ is
A) 0
B) 2
C) 1
D) 3
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Factorise: $$ 25x^{2} - 10x + 1 - 36y^{2} $$
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1 (iii) Express 3825 as the product of its prime factors.
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In the adjoining factor tree, find the numbers $\mathrm{m}$ and $\mathrm{n}$.
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How to calculate the highest 9-digit number using 8, 7, 6, 5? How to calculate similar problems using different combinations?
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