Factorize the following
$x^{2} + 15 x + 14$

(x + 1) (x + 14)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(x - 1) (x + 14)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(x + 1) (x - 14)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(x + 1) (x + 14)$
In the given polynomial $x^{2} + 15 x + 14$ ,

The first term is $x^{2}$ and its coefficient is $1$ .

The middle term is $15 x$ and its coefficient is $15$ .

The last term is a constant term $14$ .

Multiply the coefficient of the first term by the constant $1 \times 14 = 14$ .

We now find the factor of $14$ whose sum equals the coefficient of the middle term, which is $15$ and then factorize the polynomial $x^{2} + 15 x + 14$ as shown below:
$x^{2} + 15 x + 14$

$= x^{2} + x + 14 x + 14$

$= x (x + 1) + 14 (x + 1)$

$= (x + 14) (x + 1)$

Hence, $x^{2} + 15 x + 14 = (x + 14) (x + 1)$ .

Suggest Corrections

Similar questions

x - \frac{1}{x} = \frac{15}{4}

x + \frac{1}{x}

\frac{17}{4}

\frac{13}{4}

\frac{1}{4}

{tan}^{- 1} (x + \frac{2}{x})

-

{tan}^{- 1} (x - \frac{2}{x})

= {tan}^{- 1} (\frac{4}{x})

x =

If $c o s^{- 1} \frac{3}{5} - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$ then x equal to

(x - 1) {(x - 1)}^{4} {(x - 3)}^{2} (x - 4) > 0

(1, 2) \cup (2, 3) \cup (4, \infty)

t a n^{- 1} (x + \frac{2}{x}) - t a n^{- 1} (x - \frac{2}{x}) = t a n^{- 1} \frac{4}{x}

2 x^{4} - 3 x^{2}

Join BYJU'S Learning Program