what would f'x be?
f(x)= 5x^2 + 4x +3 / 2x^2 - 2x -14

i tried simplifying and using the quotient rule and got 2x-14
f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)>
equating the expression with 0,3x2 + 5x + 2 = 0on factorising it further,3x2 + 3x + 2x + 2 = 03x(x + 1) + 2(x + 1) = 0(3x + 2) (x + 1) = 0 x = - or x = -1 the zeroes of 3x2 + 5x + 2 are – and -1. let put a(x2+4)+(bx+c)(2x+1) =17put 2x+1=0a=4by equating the coefficient of x2 and constant term,a+2b=04+2b=0b=-24a+c=174×4+c=17c=1from the equation(1), we get, without calculus, the  horizontal coordinate of a quadratic of  f(x) =ax^2 + bx + c  is equal to - b/ 2a
   which in this case of
       f( x) =  - 5 x^2 + 3x + 2  
 
       equals :    - 3/[2(-5)] = -3/ -10 = 3/10 let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, hint: mathematics includes the study of topics which are related to quantity, structure, space and change. to solve this problem, we first try to simplify the expression. for doing so we let ${{x}^{2}}-4x=t$ and solve by using this consideration. in this way we can easily factorise the expression.complete step-by-step answer:according to a question, we are given the expression $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$. to simplify this complex expression, we let ${{x}^{2}}-4x=t$.this can be mathematically expressed as: $t(t-1)-20$.since, these types of factorization are familiar to us, so we try to factorise the equation in terms of t:${{t}^{2}}-t-20$now, by using the middle term splitting method of separating the variables, we get$\begin{align}  & {{t}^{2}}+4t-5t-20=0 \\  & t(t+4)-5(t+4)=0 \\  & (t-5)(t+4)=0 \\ \end{align}$ so, the factors of the expression in terms of t are (t-5) and (t+4).now, finally replacing t as ${{x}^{2}}-4x$,we get$\left( {{x}^{2}}-4x-5 \right)\left( {{x}^{2}}-4x+4 \right)=0$ on further simplifying this expression, by using middle term splitting method again in part one:$\begin{align}  & {{x}^{2}}-4x-5={{x}^{2}}+x-5x-5 \\  & \rightarrow x(x+1)-5(x+1)=0 \\  & (x-5)(x+1)=0 \\ \end{align}$ now, for the second part, using the identity of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get${{x}^{2}}-4x+4={{(x-2)}^{2}}$$\therefore (x-5)(x+1){{(x-2)}^{2}}=0$ hence, the factors of the required equation as per problem statement are: $(x-5)(x+1){{(x-2)}^{2}}=0$ note: this question can be alternatively solved by expanding the powers of x in the given expression and then letting ${{x}^{2}}$ as another variable. but the complexity of the problem would increase by doing so. hence, for simplification we must assume some other variable which can be finally replaced in the original expression to obtain the answer. in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) formula used/theory.there is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)we have,f(x) = 2 – xthen,zero of the polynomial f(x)means f(x) = 02 – x = 0-x = -2x = 2∴ zero of polynomial f(x) is 2conclusion.hence, the zero of the polynomial f(x) is 2 2x²+3x+5=0 has solutions by the quadratic formula x=(-3±√(9-4×2×5))/4 =(-3±√(-31))/4
there are no real solutions, since the discriminant is -31, whose square root is imaginary.