plzzzzzzzz sum1 help be quick n i want the full - Maths - Polynomials

Question

plzzzzzzzz sum1 help be quick n i want the full steps tht how
the answer came
1) find the values of a
and b so that the polynomial (
x4+ ax3- 7x2 +8x + b ) is exactly
divisible by (x + 2) as well as ( x+3)
2) if ( x3 + ax2 + bx + 6
) has ( x - 2 ) as a factor and leaves a remainder
3 when divided by ( x - 3 ) , find the values of
a and b

Gursheen kaur. · Accepted Answer

$1) l e t f (x) = x^{4} + a x^{3} - 7 x^{2} + 8 x + b i f x + 2 i s a f a c t o r o f f (x), t h e n f (- 2) = 0 {(- 2)}^{4} + a {(- 2)}^{3} - 7 {(- 2)}^{2} + 8 (- 2) + b = 0 16 - 8 a - 28 - 16 + b = 0 - 8 a + b = 28 8 a - b = - 28... (1) A l s o, (x + 3) i s a f a c t o r o f f (x), t h e n f (- 3) = 0 {(- 3)}^{4} + a {(- 3)}^{3} - 7 {(- 3)}^{2} + 8 (- 3) + b = 0 81 - 27 a - 63 - 24 + b = 0 - 27 a + b = 6 27 a - b = - 6..... (2) o n s u b t r a c t i n g (1) & (2), w e g e t, 8 a - 27 b = - 28 + 6 - 19 b = - 22 b = \frac{22}{19} p u t i n (1), w e g e t, 8 a - \frac{22}{19} = - 28 8 a = - 28 + \frac{22}{19} 8 a = \frac{- 510}{19} a = \frac{- 255}{76}$

$2) L e t f (x) = x^{3} + a x^{2} + b x + 6 (x - 2) i s a f a c t o r o f f (x) t h e r e f o r e, f (2) = 0 \Rightarrow 2^{3} + a 2^{2} + 2 b + 6 = 0 \Rightarrow 8 + 4 a + 2 b + 6 = 0 \Rightarrow 4 a + 2 b = - 14 \Rightarrow 2 a + b = - 7...... (1) n o w, f (x) l e a v e s r e m a i n d e r 3 w h e n d i v i d e d b y (x - 3) \Rightarrow f (3) = 3 \Rightarrow 3^{3} + 3^{2} a + 3 b + 6 = 3 \Rightarrow 27 + 9 a + 3 b + 6 = 3 \Rightarrow 9 a + 3 b = - 30 \Rightarrow 3 a + b = - 10...... (2) s u b t r a c t i n g (1) & (2), 2 a - 3 a = - 7 + 10 - a = 3 a = - 3 f r o m (1), 2 (- 3) + b = - 4 - 6 + b = - 4 b = - 4 + 6 b = 2$