If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then find the value of ‘a’ and ‘b’

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Harshitha P

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If x2 -- 3x + 2 divides x3 -- 6x2 + ax + b exactly, then the roots of x2 -- 3x + 2 will also be the roots of x3 -- 6x2 + ax + b. The roots of x2 -- 3x + 2 are 1 and 2. Therefore, substitute 1 and 2 in x3 -- 6x2 + ax + b=0. Step i: (1)3 - 6(1)2 + a (1) +b=0 1 - 6 +a+b=0 a+b=5 Step...
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If x2 – 3x + 2 divides x3 – 6x2 + ax + b exactly, then the roots of x2 – 3x + 2 will also be the roots of x3 – 6x2 + ax + b. The roots of x2 – 3x + 2 are 1 and 2. Therefore, substitute 1 and 2 in x3 – 6x2 + ax + b=0. Step i: (1)3 - 6(1)2 + a (1) +b=0 1 - 6 +a+b=0 a+b=5 Step i: (2)3 - 6(2)2 + a (2) +b=0 8 - 24 + 2a + b=0 2a+b=16 a+ (a+b)=16 a+5=16 {Since, a+b=5 in step i} a=11 b=-6 read less
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Let P(x)=x3-6x2+ax+b be given polynomial Since x2-3x+2=(x-2)(x-1) divides P(x) exactly -1,-2 are zeroes of P(x) So, p(-1)=0, P(-2)=0 We get -5+a+b=0 and -16+2a+b=0 By solving we get a=11,b=-6
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2x 3y=8 X=1 and y=8/6 X=8/6 and y=1 X=-1 and y=(-8)/6 X=(-8)/6 and y=-1
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