If the polynomial f x ax 3+bx-c is divisible
WebIf the polynomial f (x) = ax 3 + bx − c is divisible by the polynomial g (x) = x 2 + bx + c, then ab = (a) 1 (b) 1 c (c) −1 (d) - 1 c Q. If A(x) and B(x) be two polynomials and f(x)=A(x3)+xB(x3). If f(x) is divisible by x2+x+1 then show that it is divisible by x−1 … Webpls subscribe i hope it will be useful for you this question is for ntse exam use it
If the polynomial f x ax 3+bx-c is divisible
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WebThe polynomial remainder theorem follows from the theorem of Euclidean division, which, given two polynomials f(x) (the dividend) and g(x) (the divisor), asserts the existence … Web28 jul. 2024 · f ( x) is divisible by x 3. f ( x) + 2 is divisible by ( x + 1) 3. Find that polynomial. I know that because f ( x) is divisible by x 3 our polynomial is in the form …
Web12 jun. 2024 · If the polynomial f (x) = ax3 + bx – c is divisible by the polynomial g (x) = x2 + bx + c, then the value of ab is: (a) 1/c (b) 1 (c) –1 (d) none of these Answer … WebLet us see the polynomial f(x) = 2x^3 +ax^2 - bx + 3. (x+3) is one factor which means x = -3. Substitute -3 for x to get-54 +9a + 3b + 3 =0, or dividing by the HCF, which is 3 we get …
WebSoluciona tus problemas matemáticos con nuestro solucionador matemático gratuito, que incluye soluciones paso a paso. Nuestro solucionador matemático admite matemáticas básicas, pre-álgebra, álgebra, trigonometría, cálculo y mucho más. WebThe polynomial remainder theorem follows from the theorem of Euclidean division, which, given two polynomials f(x) (the dividend) and g(x) (the divisor), asserts the existence (and the uniqueness) of a quotient Q(x) and a remainder R(x) such that. If the divisor is where r is a constant, then either R(x) = 0 or its degree is zero; in both cases ...
Web24 apr. 2024 · Best answer Given; f (x) = ax3 + bx – c is divisible by g (x) = x2 + bx + c Now by division Method, As f (x) is divisible by g (x), then remainder must be 0, i.e. …
Web22 sep. 2024 · Reason: Sum of zeroes of a quadratic polynomial ax 2 +bx+c is -b/a. Answer Answer: (b) Q.9. Assertion: The graph y=f(x) is shown in figure, for the … std without sexual contactWeb8 sep. 2024 · RD Sharma Class 10 Solutions Chapter 2 MCQs (Updated for 2024-22) September 8, 2024 by Monica. RD Sharma Class 10 Solutions Chapter 2 MCQs: … std with fishy odorWebIf the zeros of the polynomial f(x)=ax 3+3bx 2+3cx+d are in A.P., then prove that 2b 3− 3abc+a 2d=0. Medium Solution Verified by Toppr Let the zeros of the given polynomial be p,q,r. As the roots are in A.P., then it can be assumed as p−k,p,p+k, where k is the common difference. p−k+p+p+k=− a3b p=− ab And, (p−k)(p)(p+k)=− ad p(p 2−k 2)=− ad std windows typefaceWebAnswer in Brief If x 3 + ax 2 − bx+ 10 is divisible by x 2 − 3x + 2, find the values of a and b Advertisement Remove all ads Solution Let f (x)= x 3 + ax 2 − bx+ 10 and g ( x) = x 2 - 3 x + 2 be the given polynomials. We have, g ( x) = x 2 - 3 x + 2 = x 2 - 2 x - x + 2 = ( x - 2) ( x - 1) Here, (x-2)and (x-1) are the factors of g(x), Now, std with dischargeWebIf the polynomial f (x) = ax 3 + bx − c is divisible by the polynomial g (x) = x 2 + bx + c, then ab = (a) 1 (b) 1 c (c) −1 (d) - 1 c Solution We have to find the value of Given is … std window testingWebTheorem 1 The polynomial interval equation F(x)+aX 2 = [L(x)+bX,U(x)+cX ] is integer solvable subject to a set of constraints on x in R and the constraint P 2(x) X Q2(x), iff the polynomial interval equation F(x)=[L(x)+(bía)X 2, U(x)+ (cía)X ], is integer solv-able subject to the same constraints. Proof Let us assume that the polynomial std with no cureWebSolutions for Chapter 4.1 Problem 80E: A cubic function is a polynomial of degree 3; that is, it has the form f (x) = ax3 + bx2 + cx + d, where a ≠ 0.(a) Show that a cubic function can … std with condom