# 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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