BeMath-Suppose-2x-3-3x-2-14x-5-Px-Q-x-3-x-1-R-for-all-value-of-x-Find-the-value-of-P-Q-and-R-

Question Number 108888 by bemath last updated on 20/Aug/20

\frac{⋮ B e M a t h ⋮}{△}

S u p p o s e 2 x^{3} + 3 x^{2} - 14 x - 5 = (P x + Q) (x + 3) (x + 1) + R f o r a l l

v a l u e o f x . F i n d t h e v a l u e o f P, Q a n d R

Answered by Rasheed.Sindhi last updated on 20/Aug/20

\underset{∙}{\overset{∙}{⟩}} \underset{∙}{\overset{∙}{∣}} \underset{∙}{\overset{∙}{⟨}} Comparing Coefficients

2 x^{3} + 3 x^{2} - 14 x - 5

= (P x + Q) (x + 3) (x + 1) + R

= {P x}^{3} + (4 P + Q) x^{2} + (3 P + 4 Q) x + 3 Q + R

C o m p a r i n g c o e f f i c i e n t s :

P = 2, 4 P + Q = 3, 3 P + 4 Q = - 14

3 Q + R = - 5

Q = 3 - 4 P = 3 - 4 (2) = - 5

Q = - 5, R = - 5 - 3 Q = - 5 - 3 (- 5)

R = 10_{⋘ ∙}^{⋙ ∙} {R a s h e e d}_{∙ ⋘}^{∙ ⋙}

Commented by bemath last updated on 20/Aug/20

c o o l l \dots n i c e

Answered by bobhans last updated on 20/Aug/20

p u t x = - 1 \to R = - 2 + 3 + 14 - 5 = 10

t h e n 2 x^{3} + 3 x^{2} - 14 x - 5 = (P x + Q) (x + 3) (x + 1) + 10

(P x + Q) (x + 3) (x + 1) = 2 x^{3} + 3 x^{2} - 14 x - 15

d i v i d e d b y x + 1 i n R H S w e g o t f a c t o r i s e

(x + 1) (2 x^{2} + x - 15) = (x + 1) (2 x - 5) (x + 3)

s o w e o b t a i n \to {\begin{cases} P = 2 \\ Q = - 5 \end{cases}

Commented by bemath last updated on 20/Aug/20

c o o l l \dots ★

Answered by Rasheed.Sindhi last updated on 20/Aug/20

\underset{\subset}{\overset{\cap}{S} y n t h e i c} Division metho \underset{\cup}{\overset{\supset}{d}}

g (x) : 2 x^{3} + 3 x^{2} - 14 x - 5

= (P x + Q) (x + 3) (x + 1) + R

g (x) o n d i v i d i n g b y x + 1 g i v e s

r e m a i n d e r R a n d q u o t i e n t q (x) :

| \begin{matrix} - 1) & 2 & 3 & - 14 & - 5 \\ - 2 & - 1 & 15 \\ 2 & 1 & - 15 & \overset{R =}{10} \end{matrix} |

q (x) = 2 x^{2} + x - 15, R = 10

q (x) o n d i v i d i n g b y x + 3 g i v e s

r e m a i n d e r 0 a n d q u o t i e n t P x + Q

| \begin{matrix} - 3) & 2 & 1 & - 15 \\ - 6 & + 15 \\ \overset{P =}{2} & \overset{Q =}{- 5} & 0 \end{matrix} |

P x + Q = 2 x - 5

P = 2, Q = - 5

Commented by Rasheed.Sindhi last updated on 20/Aug/20

B o t h s t e p s c a n b e m i x e d :

| \begin{matrix} - 1) & 2 & 3 & - 14 & - 5 \\ - 2 & - 1 & 15 \\ - 3) & 2 & 1 & - 15 & ⟨ \overset{R =}{10} ⟩ \\ - 6 & + 15 \\ \overset{P =}{2} & \overset{Q =}{- 5} & ⟨ 0 ⟩ \end{matrix} |