# Factorise:(i) $12 x^{2}-7 x+1$(ii) $2 x^{2}+7 x+3$(iii) $6 x^{2}+5 x-6$(iv) $3 x^{2}-x-4$

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To do:

We have to factorise the given expressions.

Solution:

We can find the factorisation of the given expressions by using the splitting the middle term method.

(i) $12 x^{2}-7 x+1$

We have to find two numbers whose sum is $-7$ and the product is $1\times12 = 12$

Here, $(-3)+(-4)=-7$ and $(-3)\times(-4)=12$.

Therefore,

We get $-3$ and $-4$ as the numbers.

$12x^2-7x+1=12x^2-4x-3x+1$

$=4x(3x-1)-1(3x-1)$

$=(3x-1)(4x-1)$

Hence, factors of $12x^2-7x+1$ are $(3x-1)$ and $(4x-1)$.

(ii) $2 x^{2}+7 x+3$

We have to find two numbers whose sum is $7$ and the product is $2\times3 = 6$

Here, $6+1=7$ and $6\times1=6$.

Therefore,

We get $6$ and $1$ as the numbers.

$2x^2+7x+3=2x^2+6x+x+3$

$=2x(x+3)+1(x+3)$

$=(x+3)(2x+1)$

Hence, factors of $2x^2+7x+3$ are $(x+3)$ and $(2x+1)$.
(iii) $6 x^{2}+5 x-6$

We have to find two numbers whose sum is $5$ and the product is $6\times(-6) = -36$

Here, $9+(-4)=9-4=5$ and $9\times(-4)=-36$.

Therefore,

We get $9$ and $-4$ as the numbers.

$6x^2+5x-6=6x^2+9x-4x-6$

$=3x(2x+3)-2(2x+3)$

$=(2x+3)(3x-2)$

Hence, factors of $6x^2+5x-6$ are $(2x+3)$ and $(3x-2)$.

(iv) $3 x^{2}-x-4$

We have to find two numbers whose sum is $-1$ and the product is $3\times(-4) = -12$

Here, $(-4)+3=-1$ and $(-4)\times3=-12$.

Therefore,

We get $-4$ and $3$ as the numbers.

$3x^2-x-4=3x^2-4x+3x-4$

$=3x(x+1)-4(x+1)$

$=(x+1)(3x-4)$

Hence, factors of $3x^2-x-4$ are $(x+1)$ and $(3x-4)$.

Updated on 10-Oct-2022 13:39:07