Question

12.Factorize by splitting the middle term
i) x² + 5x + 6
ii) x2 + 8x + 15

Answer 1

Answer:

i) (x + 2) + (x + 3)

ii) (x + 3) + (x + 5)

Step-by-step explanation:

Given terms :

i) x² + 5x + 6

ii) x² + 8x + 15

✦ Steps of solving :

• Multiply the first and last term with their coefficient.
• Take out the factors of the product.
• And choose a suitable factor which can split the middle term.

❐Solving for (i) :

i) x² + 5x + 6

✦Step 1 :

• Multiplication of coefficient of the first and last term

i. e., 1 × 6 = 6

✦Step 2:

Factors of 6 :

= 2*3

= 1*6

• Here we'll choose the factor (2*3) as it sums 5.

✦On Further solving :

→ x² + 5x + 6

[Splitting the middle term]

→ x² + 2x + 3x + 6

[taking commons]

→ x(x + 2) + 3(x + 2)

[taking commons]

→ (x + 2) + (x + 3)

$\therefore$ The factor of x² + 5x + 6 is :

➙ (x + 2) + (x +3)${\boxed{\green{\checkmark{}}}}$

_____________________

❐Solving for (ii) :

Following the same steps for solving :

ii) x² + 8x + 15

Can also be written as ,

→ x² + 3x + 5x + 15

[taking commons]

→ x(x + 3) + 5(x + 3)

[taking common]

→ (x + 3) + (x + 5)

$\therefore$ The factor of x² + 8x + 15 is :

➙ (x + 3) + (x + 5) ${\boxed{\green{\checkmark{}}}}$

_____________________

Answer 2

$\Huge\mathfrak\blue{answer:}$

$\large\bf{~~~~~~(~i~)~~x^{2}+5x+6}$

$\bf{Here,~~~S=5~,~~P=6~~~~(6,~2)}$

$\bf{- \longrightarrow (x^{2}+2x)+(3x+6)}$

$\bf{- \longrightarrow [x(x+2)~]+[3(x+2)~]}$

$\bf{- \longrightarrow (x+3)(x+2)}$

$\bf\therefore{The~factors~are:}$

$\bf{{\blue{x+3}}~~and~~{\blue{x+2}}}$

―――――――――――――――――――――――――――

$\large\bf{~~~~~~(~ii~)~~x^{2}+8x+15}$

$\bf{Here,~~~S=8~,~~P=15~~~(3,~5)}$

$\bf{- \longrightarrow (x^{2}+3x)+(5x+15)}$

$\bf{- \longrightarrow [x(x+3)~]+[5(x+3)~]}$

$\bf{- \longrightarrow (x+5)(x+3)}$

$\bf\therefore{The~factors~are:}$

$\bf{{\blue{x+5}}~~and~~{\blue{x+3}}}$