Question

If fx=x4-2x^3+3x^2-ax+b is divided by x-1 and x+1,it leaves remainder 5 and 19 respectively.Find the value of a and b.

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Answer 1

f(x) = 4x - 2x3 + 3x2

1. x -1=0

x=0+1

x=1

f(1) = 4(1)-2(1)3+3(1)2

f(1)= 4 - 6 + 6

f(1)= 4-12= 8

2. x+1=0

x-1= 0

x= -1

f(-1)= 4(-1) - 2(-1)3 + 3(-1)2

f(-1)= -4 - 6 + 6

f(-1)= -4+12 = 8

Answer 2

Answer:

Step-by-step explanation:

Answer:

$\begin{gathered}\begin{gathered}x + 1 = 0 \: \: \: \: , \: \: \: \: x - 1 = 0\\ x = - 1 \: \: \: \: , \: \: \: \: x = 1 \\ f(x) = {x}^{4} - 2 {x}^{3} + 3 {x}^{2} - ax + b \\ f( - 1) = {( - 1)}^{4} - 2 {( - 1)}^{3} + 3 {( - 1)}^{2} - a( - 1) + b=19 \\ = 1 + 2 + 3 + a + b =19\\ = 6 + a + b =19 \\=-13+a+b =0- (i)\\ f( 1) = {( 1)}^{4} - 2 {( 1)}^{3} + 3 {( 1)}^{2} - a( 1) + b=5 \\ = 1 - 2 + 3 - a + b=5 \\ = 2 - a + b =5\\=-3-a+b=0 - (ii)\\ \blue{\mathfrak{\underline{\large{By \: {eq}^{n} \: (i) \: and \: (ii) }}}:}\\ -13 + a + b = -3 - a + b \\ 2a = 10 \\ \boxed{a = 5} \\ \blue{\mathfrak{\underline{\large{ put \: the \: value \: of \: ‘a’ \: in \: (i)}}}:} \\ -13 +5+ b = 0 \\ \boxed{b = 8} \\ \blue{\mathfrak{\underline{\large{Verifying}}}:} \\ f( - 1) = -13 + a + b \\ = -13 +5 +8 = 0\\f( 1) = -3 -a+b \\ = -3-5 +8 = 0 \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}\end{gathered} \end{gathered} </p><p></p><p>$