These Class 8 Maths Extra Questions Chapter 14 Factorization will help students prepare well for the exams.
Class 8 Maths Chapter 14 Extra Questions Factorization
Factorization Extra Questions Class 8
Question 1.
Factorise 3x2y – 15xy.
Solution:
3×2 – 15xy
= 3(x)(x)(y) – 3(5)(x)(y)
= 3xy[x – 5]
∴ 3x2y – 15xy = 3xy[x – 5].
Question 2.
Factorize the following expressions.
a) 3ax – 6xy + 8by – 4ba
b) 2xy + 8ky + mx + 4km
Solution:
a) 3ax – 6xy + 8by – 4ba
= 3x(a – 2y) + 4b(2y – a)
= 3x(a – 2y) – 4b(a – 2y)
= (a – 2y) (3x – 4 b)
∴ 3ax – 6xy + 8by – 4ba = (a – 2y) (3x – 4b)
Factors of 3ax – 6xy + 8by – 4ba are (a – 2y) and (3x – 4b)
b) 2xy + 8ky + mx + 4km
= 2y(x + 4k) + m(x + 4k)
= (x + 4k) (2y + m)
2xy + 8ky + mx + 4km = (x + 4k) (2y + m)
∴ Factors of 2xy + 8ky + mx + 4km are (x + 4k) and (2y + m).
Factorization Class 8 Extra Questions
Question 1.
Factorise 12a2b + 15ab2
Solution:
We have 12a2b = 2 × 2 × 3 × a × a v b
15ab2 = 3 × 5 × a × b × b
The two terms have 3, a and b as common factors.
Therefore,
12a2b + 15ab2 = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
= 3 × a × b × [(2 × 2 × a) + (5 × b)]
= 3ab × (4a + 5b)
= 3ab (4a + 5b) (required factor form)
Question 2.
Factorise 10x2 – 18x3 + 14x 4
Solution:
10x2 = 2 × 5 × x × x
18x3 = 2 × 3 × 3 × x × x × x
14x4 = 2 × 7 × x × x × x × x
The common factors of the three terms are 2, x and x.
Therefore,
10x2 – 18x3 + 14x4
= (2 × x × x × 5) – (2 × x × x × 3 × 3 × x) + (2 × x × x × 7 × x × x)
= 2 × x × x × [(5 – (3 × 3 × x) + (7 × x × x)]
= 2x2 × (5 – 9x + 7x2)
= 2x2(7x2 – 9x + 5)
Question 3.
Factorise 6xy – 4y + 6 – 9x.
Solution:
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor 2y;
6xy – 4y = 2y (3x – 2) ________ (a)
Observe them. If you change their order to – 9x + 6, the factor (3x – 2) will come out;
– 9x + 6 = -3 (3x) + 3 (2) = – 3 (3x – 2) ________ (b)
Step 3 Putting (a) and (b) together, 6xy – 4y + 6 – 9x
= 6xy – 4y – 9x + 6
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)
The factors of (6xy – 4y + 6 – 9x) are (3x – 2) and (2y – 3).
Question 4.
Factorise x2 + 8x + 16
Solution:
Observe the expression; it has three terms. Therefore, it does not fit Identity III. Also, it’s first and third terms are perfect squares with a positive sign before the middle term.
So, it is of the form a2 + 2ab + b2
where a = x and b = 4
such that
a2 + 2ab + b2 = x2 + 2 (x) (4) + 42
= x2 + 8x + 16
Since a2 + 2ab + b2 = (a + b)2,
by comparison x2 + 8x + 16 = (x + 4)2 (the required factorisation)
Question 5.
Factorise 4y2 – 12y + 9.
Solution:
Observe 4y2 = (2y)2,
9 = 32 and 12y = 2 × 3 × (2y)
Therefore, 4y2 – 12y + 9
= (2y)2 – 2 × 3 × (2y) + (3)2
(Observe here the given expression is of the form a2 – 2ab + b2. Where a = 2y, and b = 3 with 2ab = 2 × 2y × 3 = 12y.)
= (2y – 3)2 (required factorisation)
Question 6.
Factorise 49p2 – 36
Solution:
There are two terms; both are squares and the second is negative. The expression is of the form (a2 – b2). Identity III is applicable here;
49p2 – 36 = (7p)2 – (6)2
= (7p – 6) (7p + 6) (required factorisation)
Question 7.
Factorise a2 – 2ab + b2 – c2
Solution:
The first three terms of the given expression form (a – b)2. The fourth term is a square. So the expression can be red¬uced to a difference of two squares.
Thus, a2 – 2ab + b2 – c2
= (a – b)2 – c2 (Applying Identity II)
= [(a – b) – c) ((a – b) + c)] (Applying Identity III)
= (a – b – c) (a – b + c) (required factorisation)
Notice, how we applied two identities one after the other to obtain the required factorisation.
Question 8.
Factorise m4 – 256
Solution:
We note m4 = (m2)2 and 256 = (16)2
Thus, the given expression fits Identity III.
Therefore, m4 – 256
= (m2)2 – (16)2
= (m2 – 16) (m2 + 16) [(using Identity (III)]
Now, (m2 + 16) cannot be factorised further, but (m2 – 16) is factorisable again as per Identity III.
m2 – 16 = m2 – 42 = (m – 4) (m + 4)
Therefore,
m4 – 256 = (m – 4) (m + 4) (m2 + 16)
Question 9.
Factorise x2 + 5x + 6
Solution:
If we compare the R.H.S. of Identity (IV) with x2 + 5x + 6,
we find ab = 6, and a + b = 5.
From this, we must obtain a and b. The factors then will be (x + a) and (x + b).
If ab = 6, it means that a and b are factors of 6.
Let us try a = 6, b = 1.
For these values a + b = 7, and not 5, So this choice is not right.
Let us try a = 2, b = 3. For this a + b = 5 exactly as required.
The factorised form of this given expre-ssion is then (x + 2) (x + 3).
Question 10.
Find the factory of y2 – 7y + 12.
Solution:
We note 12 = 3 × 4 and 3 + 4 = 7.
Therefore,
y2 – 7y + 12 = y2 – 3y – 4y + 12
= y (y – 3) – 4 (y – 3)
= (y – 3) (y – 4)
Question 11.
Obtain the factors of z2 – 4z – 12.
Solution:
Here a b = -12; this means one of a and b is negative. Further, a + b = – 4, this means the one with larger numerical value is negative. We try a = -4, b = 3; but this will not work, since a + b = -1. Next possible values are a = -6, b = 2, so that a + b = -4 as required.
Hence, z2 – 4z – 12 = z2 – 6z + 2z -12
= z(z – 6) + 2(z – 6)
= (z – 6) (z + 2)
Question 12.
Find the factors of 3m2 + 9m + 6.
Solution:
We notice that 3 is a common factor of all the terms.
Therefore, 3m2 + 9m + 6 = 3(m2 + 3m + 2)
Now, m2 + 3m + 2 = m2 + m + 2m + 2 (as 2 = 1 × 2)
= m(m + 1)+ 2(m + 1)
= (m + 1) (m + 2)
Therefore
3m2 + 9m + 6 = 3(m + 1) (m + 2)
For factorising an algebraic expression in the form of, x2 + ax + b;
procedure as follows
we find factors p, q of ’b’ such that
p + q = a and pq = b
then we write given x2 + ax + b as follow
= x2 + px + qx + pq
= x(x + p) + q(x + p)
= (x + p) (x + q)
thus we factorise given x2 + ax + b as (x + p)(x + q) such that pq = b, p + q = a
Question 13.
Do the following divisions.
i) -20x4 ÷ 10x2
ii) 7x2y2z2 ÷ 14xyz
Solution:
i) -20x4 = -2 × 2 × 5 × x × x × x × x
10x2 = 2 × 5 × x × x
Therefore, (-20x4) ÷ 10x2
= \(\frac{-2 \times 2 \times 5 \times \mathrm{x} \times \mathrm{x} \times \mathrm{x} \times \mathrm{x}}{2 \times 5 \times \mathrm{x} \times \mathrm{x}}\)
= -2 × x × x = -2x2
ii) 7x2y2z2 ÷ 14xyz
= \(\frac{7 \times \mathrm{x} \times \mathrm{x} \times \mathrm{y} \times \mathrm{y} \times \mathrm{z} \times \mathrm{z}}{2 \times 7 \times \mathrm{x} \times \mathrm{y} \times \mathrm{z}}\)
= \(\frac{\mathrm{x} \times \mathrm{y} \times \mathrm{z}}{2}\) = \(\frac{1}{2}\)xyz
Question 14.
Divide 24(x2yz + xy2z + xyz2) by 8xyz using both the methods.
Solution:
24 (x2yz + xy2z + xyz2)
= 2 × 2 × 2 × 3 × [(x × x × y × z) + (x × y × y × z) + (x × y × z × z)]
= 2 × 2 × 2 × 3 × x × y × z ×(x + y + z)
(by taking out the common factor)
= 8 × 3 × xyz × (x + y + z)
Therefore, 24 (x2yz + xy2z + xyz2) ÷ 8xyz
= \(\frac{8 \times 3 \times \mathrm{xyz} \times(\mathrm{x}+\mathrm{y}+\mathrm{z})}{8 \times \mathrm{xyz}}\)
= 3 × (x + y + z) = 3 (x + y + z)
Alternately, 24(x2yz + xy2z + xyz2) ÷ 8xyz
= \(\frac{24 x^2 y z}{8 x y z}\) + \(\frac{24 x y^2 z}{8 x y z}\) + \(\frac{24 \mathrm{xyz}^2}{8 \mathrm{xyz}}\)
= 3x + 3y + 3z = 3(x + y + z)
Question 15.
Divide 44(x4 – 5x3 – 24x2) by 11x (x – 8)
Solution:
Factorising 44(x4 – 5x3 – 24x2), we get 44(x4 – 5x3 – 24x2)
= 2 × 2 × 11 × x2(x2 – 5x – 24)
(taking the common factor x2 out of the bracket)
= 2 × 2 × 11 × x2(x2 – 8x + 3x – 24)
= 2 × 2 × 11 × x2 [x (x – 8) + 3(x – 8)]
= 2 × 2 × 11 × x2 (x + 3) (x – 8)
Therefore,
44(x4 – 5x3 – 24x2) ÷ 11x(x – 8)
= \(\frac{2 \times 2 \times 11 \times \mathrm{x} \times \mathrm{x} \times(\mathrm{x}+3) \times(\mathrm{x}-8)}{11 \times \mathrm{x} \times(\mathrm{x}-8)}\)
= 2 × 2 × x (x + 3) = 4x(x + 3)
(We cancel the factors 11, x and (x – 8) common to both the numerator and denominator.)
Question 16.
Divide z(5z2 – 80) by 5z(z + 4)
Solution:
Dividend = z(5z2 – 80)
= z[(5 × z2) – (5 × 16)]
= z × 5 × (z2 – 16)
– 5z × (z + 4) (z – 4) [using the identity a2 – b2 = (a + b) (a – b)]
Thus, z(5z2 – 80) ÷ 5z(z + 4)
= \(\frac{5 z(z-4)(z+4)}{5 z(z+4)}\) = (z – 4)