Question

Factorise \[15xy - 6x + 5y - 2\]
a) \[(x - 1)(y + 2)\]
b) \[(x + 1)(y - 2)\]
c) \[(3x + 1)(5y - 2)\]
d) \[(3x - 1)(5y + 2)\]

Answer 1

Hint: We need to factorise the given expression. For factoring, we will first see which of the terms can be decomposed into factors. Here we can decompose \[15xy\] and \[6x\] as \[(5 \times 3)xy\] and \[(2 \times 3)x\] respectively. After writing \[15xy\] in simplified factors, we now assign \[x\] and \[y\] with \[3\] and \[5\] respectively. We did this so we can take some terms common from the first two terms and some from the next two terms and then we will have two terms in the expression, out of which we will take some part common and then arrive at one of the options given in the question.

Complete step by step solution:
Factorising \[15xy - 6x + 5y - 2\] -------(1)
We will first decompose \[15xy\] into factors.
 \[15xy\] can be written as \[15xy = (3 \times 5)xy\]
 \[15xy = 3 \times 5 \times x \times y\]
After Rearranging the terms as multiplication is associative, we get
 \[15xy = 3 \times x \times 5 \times y\]
Combining first two and last two terms together
 \[15xy = 3x \times 5y\]
We assigned in such a way because we see that we have a similar terms in the remaining expression i.e. \[5y\]
Now, writing \[15xy\] as \[(3x \times 5y)\] in (1), we get
 \[15xy - 6x + 5y - 2 = (3x \times 5y) - 6x + 5y - 2\]
Now, Separating the first two terms together and next two together
 \[ = [(3x \times 5y) - 6x] + [5y - 2] \] -----(2)
Now, we can \[6x\] as \[(2 \times 3)x\] which is further equal to \[(2 \times 3x)\] .
Writing \[6x = (2 \times 3x)\] in (2), we get
 \[15xy - 6x + 5y - 2 = [(3x \times 5y) - (2 \times 3x)] + [5y - 2] \]
Now, Using Distributive Property of Multiplication i.e. \[a(b \pm c) = ab \pm ac\]
 \[15xy - 6x + 5y - 2 = [3x(5y - 2)] + [5y - 2] \] -----(3)
 \[[5y - 2] \] can be written as \[[1(5y - 2)] \] as \[1\left( a \right) = a\] ----(4)
Using (4), in (3), we get
 \[15xy - 6x + 5y - 2 = [3x(5y - 2)] + [1(5y - 2)] \]
Removing the separation Brackets
 \[15xy - 6x + 5y - 2 = 3x(5y - 2) + 1(5y - 2)\]
Now, Taking \[(5y - 2)\] as common factor from both the terms or we can say, using Distributive Property, we have
 \[15xy - 6x + 5y - 2 = (5y - 2)(3x + 1)\] ( \[a(b + c) = ab + ac\] )
Hence, we get \[15xy - 6x + 5y - 2 = (5y - 2)(3x + 1) = (3x + 1)(5y - 2)\] (Rearranging terms)
 \[\therefore \] The correct option is (c)
So, the correct answer is “Option c”.

Note: For factorisation, we need to be very careful while we are arranging the terms and decomposing the terms. We have to decompose the terms in such a manner so that we can get something common from the remaining terms and the decomposed term. We could have written \[15xy\] as \[(3y \times 5x)\] but that would be of no use as we won’t get any common meaning from the remaining terms. And, while rearranging, we need to see which terms can be arranged with which term so that some of the terms can be taken out as common terms.