factorises the given expression: $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$.
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.
solution: ` x = {:((-4/5,-1),(2/5,1)):}` formula :- therefore ,put -2x dx = dt 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] 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

:)> Can you show the calculate function that you've wrote so far? 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, 1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 } find and correct the errors in the following mathematical statements

q19) applying and factorization, we get, also changing by reversing n and dnote: - try to factorize that term which help in reducing expression. 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.