Question

If x2 +bx+b is a factor of x3+2x2+2x+c find b-c

Answer 1

Step-by-step explanation:

check the solution of the problem

Answer 2

$Let \: the \: polynomial :\\p(x) = x^{3}+2x^{2}+2x+c$

$and \: g(x) = x^{2} + bx + b \: is \: a \: factor .$

$Form \: the \: attachment \: above , [/tex</p><p>Dividend = p(x) ,</p><p>Divisor = g(x) ,</p><p>Quotient q(x) = x + (2-b)</p><p>Remainder r(x) = [(2-b)-(2-b)b]x + c-b(2-b)</p><p>But <strong>x²</strong><strong>+</strong><strong>bx</strong><strong>+</strong><strong>b</strong><strong> </strong><strong>is</strong><strong> </strong><strong>a</strong><strong> </strong><strong>fa</strong><strong>ctor</strong><strong> </strong><strong>(</strong><strong> </strong><strong>g</strong><strong>iven</strong><strong> </strong><strong>)</strong></p><p>[tex] \therecore i) (2-b)-(2-b)b = 0 \\\implies (2-b)(1-b) = 0 \\\implies (2-b) = 0 \:Or \: 1-b = 0$

$\implies \pink { b = 2 \: Or \: b = 1}$

$2. c - b(2-b) = 0 \\\implies c = b(2-b) \: --(1)$

Case 1:

$Put \: b = 2 \: then c = 2(2-2)$

$\implies c = 0$

Case 2:

$Put \: b = 1 \: then c = 1(2-1)$

$\implies c = 1$

$Therefore ( b , c ) = (2,0) \: Or \: ( 1 , 1 )$

$Now, \red { Value \: of \: b - c } \\= 2 - 0 \\= 2$

$Or, \red { Value \: of \: b - c } \\= 1 - 1 \\= 0$

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