Question

if x+1 is a factor of x^4+ax^3+2bx^2+3x=4,find a and b. given that a+4b =12

Answer 1

Taking  $p(x)=x^4+ax^3+2bx^2+3x+4$,

If  $(x+1)$  is a factor of  $p(x)=x^4+ax^3+2bx^2+3x+4$, then  $p(-1)=0$

So,

$p(-1)=(-1)^4+a(-1)^3+2b(-1)^2+3(-1)+4=0\\ \\ p(-1)=1-a+2b-3+4=0\\ \\ p(-1)=2b-a+2=0\\ \\ \\ \\ \begin{aligned}2b-a+2&=0\\ \\ 2b-a&=-2\\ \\ a-2b&=2\\ \\ a-2b+6b&=2+6b\\ \\ a+4b&=2+6b\\ \\ 12&=2+6b\ \ \ \ \ \ [\ \because\ a+4b=12\ \ \ \ \ \longrightarrow\ \ \ (1)\ ]\\ \\ 12-2&=6b\\ \\ 10&=6b\\ \\ \large \text{ \bold{\dfrac{5}{3}} }&\ \large \bold{=b}\end{aligned}$

From (1),

$\begin{aligned}a+4b&=12\\ \\ a+4\left(\dfrac{5}{3}\right)&=12\\ \\ a+\dfrac{20}{3}&=\dfrac{36}{3}\\ \\ a&=\dfrac{36}{3}-\dfrac{20}{3}\\ \\ \large \bold{a}&\large \text{ \bold{\ =\dfrac{16}{3}} }\end{aligned}$

Hence,

$\large \boxed{\boxed{\bold{a=\dfrac{16}{3}}}} \quad \quad \quad \boxed{\boxed{\bold{b=\dfrac{5}{3}}}}$

Answer 2

$\Large{\sf{\underline{\underline{Solution:}}}} \\ \\ </p><p></p><p> \large\sf Given \colon \\ \sf p(x ) = {x}^{4} + a {x}^{3} + 2b {x}^{2} + 3x + 4 \\ \sf a + 4b = 12 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(i) \\ \\ \large \sf T o \: F ind \colon \\ \sf V alue \: of \: \rm{a} \: and \: \rm{b}. \\ \\ \\ \sf x+1 \: is \: a \: factor \: of \: x^4+ax^3+2bx^2+3x + 4. \\ \sf p( - 1) = 0 \\ \\ \sf p( - 1) = ( - 1)^4+a( - 1)^3+2b( - 1)^2+3 \times -1 + 4 = 0 \\ \: \: \: \: \: \: \: \: \implies \sf 1 - 1a + 2b - 3 + 4 = 0 \\ \: \: \: \: \: \: \: \: \implies \sf 2 + 2b - a = 0 \\ \: \: \: \: \: \: \: \: \implies \sf 2b - a = - 2 \\ \: \: \: \: \: \: \: \: \implies \sf a - 2b = 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf( A dding \: 6b\: to \: both \: side )\\ \: \: \: \: \: \: \: \: \implies \sf a - 2b + 6b = 2 + 6b \\ \: \: \: \: \: \: \: \: \implies \sf a + 4b = 2 + 6b \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( \because a + 4b = 12 \: \: \: \: \: \: ....(i) )\\ \: \: \: \: \: \: \: \: \implies \sf 12 = 2 + 6b \\ \: \: \: \: \: \: \: \: \implies \sf 12 - 2 = 6b \\ \: \: \: \: \: \: \: \: \implies \sf \frac{10}{6} = b \\ \: \: \: \: \: \: \: \: \implies \boxed{ \bf \large \frac{5}{3} = b } \\ \\ \sf From \: Eq_n (i), \: We \: have \\⇒ \sf a + 4b = 12 \\ ⇒ \sf a + 4 \times \frac{5}{3} = 12 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \because \sf b = \frac{5}{3} \bigg) \\ ⇒ \sf a + \frac{20}{3} = 12 \\ ⇒ \sf a = 12 - \frac{20}{3} \\ ⇒ \boxed{\bf \large a = \frac{16}{3} }$