Question

Q.6. The values of p and q if (x+2) is a factor of f(x) =x3 + px2 + qx+6
and f(1) =24 are
a) p = 3, q =5
b) p = 6, q = 11
c) p = 5, q =6
d) p = -2, q =7

Answer 1

Answer :

Option (b) p = 6, q = 11 is the correct option.

Step by Step Explanation :

As we are given that,

(x + 2) is a factor of f(x) = x³ + px² + qx + 6

Therefore x = -2 will be a zero of f(x) which means :

$\Large \bigstar \large \: \: \: \underline{ \boxed{ \bf f( - 2) = 0 }}$

From above we get ;

$\\ \rm ( - 2) {}^{3} + p{( - 2)}^{2} + q( - 2) + 6 = 0$

$:\longmapsto \rm - 8 + 4p - 2q + 6 = 0$

$:\longmapsto \rm 4p - 2q - 2 = 0$

∆ Dividing whole equation by 2 ;

$\\ \rm2p - q - 1 = 0$

$\green{:\longmapsto \bf \underline{\underline{ \blue{ 2p - q = 1}}}} \: \: - - - - (1)$

Also we have given that :

$\Large \bigstar \large \: \: \: \underline{ \boxed{ \bf f(1) = 24}}$

From above we get ;

$\\ \rm(1 {)}^{3} + p {(1)}^{2} + q(1) + 6 = 24$

$:\longmapsto \rm 1 + p + q + 6 = 24$

$:\longmapsto \rm p + q + 7 = 24$

$:\longmapsto \rm p + q = 24 - 7$

$\green{:\longmapsto \bf \underline{\underline{ \blue{ p + q = 17}}}} \: \: - - - - (2)$

➡️ Adding (1) and (2) :

$\\:\longmapsto \rm2p - q + p + q = 1 + 17$

$:\longmapsto \rm 3p = 18$

$:\longmapsto \rm p = \cancel\frac{18}{3}$

$\large \green{:\longmapsto \bf \underbrace{ \underline{\underline{ \blue{ p = 6}}}}}$

➡️ Putting value of p in (2) :

$\\ :\longmapsto \rm 6+ q = 17$

$:\longmapsto \rm q = 17 - 6$

$\large \green{:\longmapsto \bf \underbrace{ \underline{\underline{ \blue{ q = 11}}}}}$

Answer 2

$\Large\:{\bf{\dag}\:{\underline{\underline{\frak{AnswEr\::}}}}}$

Given that: (x + 2) is a factor of f(x) = x³ + px² + qx + 6 and f(1) = 24.

Need to find: The values of p and q?

Let's do it!!

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━

✇ As we know that (x + 2) is a factor of f(x) = x³ + px² + qx + 6. Therefore;

➧ x + 2 = 0

➧ x = 0 - 2

➧ x = -2

So,

➧ f(-2) = (-2)³ + p(-2)² + q(-2) + 6 = 0

➧ -8 + 4p - 2q + 6 = 0

➧ 4p - 2q - 8 + 6 = 0

➧ 4p - 2q - 2 = 0

➧ 2(2p - q - 1) = 0

➧ 2p - q - 1 = 0/2

➧ 2p - q - 1 = 0 ⠀⠀⠀⠀— ①

✇ Again we know that f(1) = 24. Therefore;

➧ f(1) = (1)³ + p(1)² + q(1) + 6 = 24

➧ 1 + 1p + 1q + 6 = 24

➧ p + q + 1 + 6 = 24

➧ p + q + 7 = 24

➧ p + q + 7 - 24 = 0

➧ p + q - 17 = 0 ⠀⠀⠀⠀— ②

~ By adding ① and ② we get ::

➧ (2p - q - 1) + (p + q - 17) = 0 + 0

➧ 2p - q - 1 + p + q - 17 = 0

➧ 2p + p - q + q - 1 - 17 = 0

➧ 3p - 18 = 0

➧ 3p = 0 + 18

➧ 3p = 18

➧ p = 18/3

➧ p = 6

~ Putting p = 6 in ① ::

➧ 2(6) - q - 1 = 0

➧ (2 × 6) - q - 1 = 0

➧ 12 - q - 1 = 0

➧ -q - 1 + 12 = 0

➧ -q + 11 = 0

➧ -q = 0 - 11

➧ -q = -11

➧ q = 11

∴ The values of p and q is 6 and 11.

$\rule{200}7$