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Given f(x)=(x-a)^(2)g(x) , show that f^(')(x) is divisible by (x-a) . (03 marks ) A polynomial P(x)=x^(3)+4ax^(2)+bx+3 is divisible by (x-1)^(2) Use the result in (a) above to find the values of a and b . Hence solve the equation P(x)=0 (09 marks)

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Given f(x)=(x-a)^(2)g(x) , show that f^(')(x) is divisible by (x-a) . (03 marks ) A polynomial P(x)=x^(3)+4ax^(2)+bx+3 is divisible by (x-1)^(2) Use the result in (a) above to find the values of a and b . Hence solve the equation P(x)=0 (09 marks)

Given f(x)=(x-a)^(2)g(x) , show that f^(')(x) is divisible by (x-a) . (03 marks ) A polynomial P(x)=x^(3)+4ax^(2)+bx+3 is divisible by (x-1)^(2) Use the result in (a) above to find the values of a and b . Hence solve the equation P(x)=0 (09 marks)

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PatrickProfessional · Tutor for 6 years

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<p> 1. For the first part of the question, f′(x)=(x−a)2 g ′(x)+2(x−a)g(x) <br />2. For the second part, a = -6, b = 3.<br />3. For the equation P(x)=0, x = 1.</p>

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<p> <br />1. According to the product rule of differential, which states that the derivative of two-multiplicative functions is the first function times the derivative of the second function plus the derivative of the first function times the second function. In this case, if f(x)=(x−a)2 g(x), By applying the product rule, f′(x) ⇒ ((x−a)2)' - g(x) + (x-a)2 × g ′(x)⇒ (x−a)2 g ′(x)+2(x−a) g(x). Since (x–a) is a common term, it stably concludes that f′(x) must be divisible by (x−a).<br /><br />2. For the polynomial P(x)=x^3+4ax^2+bx+3 being divisible by (x-1)^2 it should at x = 1, equal 0; As per the Factor Theorem: P(x=1) = 0 ⇒ 0 = (1)^3 + 4a(1)^2 + b*1 + 3 = 1+4a+b+3 .<br />Moving terms around, then gives :4a+b+4 = 0. Further since our function gives that it's derivative is equal to 0 for a value x. g'(x=1) = 0; we had f(x) = x^3+4ax^2+bx+3 - g(x), thus f´(x) = 0 ; x=1. So, calculating f´(1) with values known, 0 = 3*(1)^2 +8a*1 + b = 3 + 8a + b, hence 8a + b = -3. From the 2 above quick values above 4a+b → Is a constant line Simplified form : b = -4a; . Putting b = 4a into 8a puntos we see that a = -6, which further lead to introducing the values b = 4a into 8a pontos and substitutating a=-6 calculated beforehand gives b = 3.<br /><br />3. Once calculated a and b, substituting them into the original Function : P(x) = x³ - 24*x² + 3*x +3, hence by applying with numerical methods x ↦ x - P(x)/P´(x) multiple time until infinitesimale changes conclude stable stopping criteria. which concludes that the there is only one answer to the equation P(x) =0, 3x = 1 </p>
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