MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Factorisation by Common Factors

Factorise:1 p...

Question

Factorise:

(1) p² − 2p − 168
(2) 6x² − 19x − 7
(3) (3x + 2)³ − 8
(4) (2x − 3)³ + 125

Open in App

Solution

(1) p² − 2p − 168 can be factorised as:
$p^{2} - 2 p - 168 = p^{2} - 14 p + 12 p - 168 = p (p - 14) + 12 (p - 14) = (p - 14) (p + 12)$

(2) 6x² − 19x − 7 can be factorised as:
$6 x^{2} - 19 x - 7 = 6 x^{2} - 21 x + 2 x - 7 = 3 x (2 x - 7) + 1 (2 x - 7) = (3 x + 1) (2 x - 7)$

(3) (3x + 2)³ − 8 can be factorised as:
${(3 x + 2)}^{3} - 8 = {(3 x + 2)}^{3} - {(2)}^{3} [a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)] = (3 x + 2 - 2) ({(3 x + 2)}^{2} + {(2)}^{2} + 2 (3 x + 2)) = 3 x (9 x^{2} + 4 + 12 x + 4 + 6 x + 4) = 3 x (9 x^{2} + 18 x + 12) = 9 x (3 x^{2} + 6 x + 4)$

(4) (2x − 3)³ + 125 can be factorised as:
${(2 x - 3)}^{3} + 125 = {(2 x - 3)}^{3} + {(5)}^{3} [a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)] = (2 x - 3 + 5) [{(2 x - 3)}^{2} + 5^{2} - (2 x - 3) 5] = (2 x + 2) [4 x^{2} + 9 - 12 x + 25 - (2 x - 3) (5)] = 2 (x + 1) [4 x^{2} + 9 - 12 x + 25 - 10 x + 15] = 2 (x + 1) [4 x^{2} + - 22 x + 49]$

Suggest Corrections

thumbs-up

0

Similar questions

Q. f(x) = x³ −6x² − 19x + 84, g(x) = x − 7

Q. Question 23
Factorise
(i)

x^{2} + 9 x + 18

6 x^{2} + 7 x - 3

2 x^{2} - 7 x - 15

84 - 2 r - 2 r^{2}

Q. Using the factor theorem factorise the following polynomial completely

3 x^{3} + 2 x^{2} - 19 x + 6

p^{2} - 2 p - 35

Show that $(x + 4), (x - 3)$ and $(x - 7)$ are factors of $x^{3} - 6 x^{2} - 19 x + 84$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Factorisation by Common Factors

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Factorisation by Common Factors

Standard VIII Mathematics

Join BYJU'S Learning Program

CrossIcon