Question

3. Factor by spilliting middle term:
(1)
$x2 - 11x - 42$
​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ x² - 11x - 42 = 0.

As we know that,

Factorizes the equation into middle term splits, we get.

⇒ x² - 14x + 3x - 42 = 0.

⇒ x(x - 14) + 3(x - 14) = 0.

⇒ (x + 3)(x - 14) = 0.

⇒ x = - 3  and  x = 14.

                                                                                                                       

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\: {x}^{2} - 11x - 42$

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers h and k such that h + k = b and hk = ac.

After finding h and k, we split the middle term in the quadratic as hx + kx and get desired factors by grouping the terms.

Now,

According to given expression,

hk = - 42 and h + k = - 11

So, we have to find h and k in such a way that h + k = - 42 and hk = - 11

We know, -14 × 3 = - 42 and - 14 + 3 = - 11

Hence,

$\rm :\longmapsto\: {x}^{2} - 11x - 42$

can be factorized using middle term splitting as

$\rm \:  =  \:  \: {x}^{2} - 14x + 3x - 42$

$\rm \:  =  \:  \: x(x - 14) + 3(x - 14)$

$\rm \:  =  \:  \: (x - 14)(x + 3)$

Thus,

$\rm :\longmapsto\: {x}^{2} - 11x - 42 = (x - 14)(x + 3)$