Ex.14.1 Q3 Factorization - NCERT Maths Class 8

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Question

Factorize:

(i) \(\begin{align} {x^2} + xy + 8x + 8y\end{align}\)

(ii) \(\begin{align} 15xy - 6x + 5y - 2\end{align}\)

(iii) \(\begin{align}ax + bx - ay - by\end{align}\)

(iv) \(\begin{align} 15pq + 15 + 9q + 25p\end{align}\)

(v) \(\begin{align}z - 7 + 7xy - xyz\end{align}\)

 Video Solution
Factorisation
Ex 14.1 | Question 3

Text Solution

What is known:

Algebraic expression.

What is unknown:

Factorisation of given algebraic expression.

Reasoning:

There are \(4\) terms in each expression. First we will make pair of two terms from which we can take out common factors and convert the expression of \(4\) terms into \(2\) terms expression then take out common factors from remaining \(2\) terms.

Steps:

\(\begin{align}({\rm{i}}) \quad & {x^2} + xy + 8x + 8y \\ &= \begin{Bmatrix}  x \times x + x \times y \\ + 8 \times x + 8 \times y \end{Bmatrix} \\&= x(x + y) + 8(x + y)\\&= (x + y)(x + 8)\end{align}\)

\(\begin{align}{\rm{(ii)}} \quad & 15xy - 6x + 5y - 2 \\ &= \begin{Bmatrix} 3 \times 5 \times x \times y - 3 \times 2 \times x +\\  5 \times y - 2 \end{Bmatrix} \\&= 3x(5y - 2) + 1(5y - 2)\\ &= (5y - 2)(3x + 1)\end{align}\)

\(\begin{align}{\rm{ (iii)}} \quad & ax + bx - ay - by \\ &=  a \times x + b  \times x - a \times y  - b \times y \\ &= x(a + b) - y(a + b)\\ &= (a + b)(x - y)\end{align}\)

\(\begin{align}{\rm{ (iv)}} \quad &15pq + 15 + 9q + 25p \\ &= 15pq + 9q + 25p + 15\\ &= \begin{Bmatrix} 3 \times 5 \times p \times q  + 3 \times 3 \times q + \\5 \times 5  \times p + 3 \times 5 \end{Bmatrix} \\&= 3q(5p + 3) + 5(5p + 3)\\&= (5p + 3)(3q + 5)\end{align}\)

\(\begin{align}({\rm{v}}) \quad & z - 7 + 7xy - xyz \\ &= \begin{Bmatrix} z - x \times y  \times z - \\7 + 7  \times x \times y \end{Bmatrix} \\&= z(1 - xy) - 7(1 - xy)\\&= (1 - xy)(z - 7)\end{align}\)

  
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