If f(x) is a polynomial such that f(1) =1, f(2) =2, f(3) =3 and f(4) =16. Find the value of f(5)?
Let us assume : [math]f(x) = ax^3 + bx^2 + cx + d [/math]Then we get, [math]f(1) = a + b + c + d = 1 ... (1)[/math][math]f(2) = 8a + 4b + 2c + d = 2 ... (2)[/math][math]f(3) = 27a + 9b + 3c + d = 3 ...(3)[/math][math]f (4) = 64a + 16b + 4c + d = 16 ...(4)[/math]Solving (1) and (2), We get[math]7a + 3b + c = 1 ...(5)[/math]Solving (1) and (3), We get[math]26a + 8b + 2c = 2 ...(6)[/math]Solving (1) and (4), We get[math]63a + 15b + 3c = 15 ...(7)[/math]Solving (5) and (6)[math]6a + b = 0[/math]Solving (5) and (7)[math]7a + b = 2[/math]This gives, a = 2, b = -12And then, c = 23 and d = - 12Thus,[math] f(x) = 2x^3 - 12x^2 + 23x - 12[/math]And So, [math]f(5) = 2*(5)^3 - 12*(5)^2 + 23(5) - 12 = 53[/math]The required answer is f(5) = 53
The polynomial p(x) =x^4-2x^3+3x^2-ax+b when divided by (x-1) and (x+1) leaves the remainders 5 and 19, respectively. What are a and b?
We have,p(x)=x^4–2x^3+3x^2-ax+bBy remainder theorem, when p(x) is divided by (x-1) and (x+1) , the remainders are equal to p(1) and p(-1) respectively.By the given condition, we havep(1)=5 and p(-1)=19=> (1)^4–2(1)^3+3(1)^2-a(1)+b=5 and (-1)^4–2(-1)^3+3(-1)^2-a(-1)+b=19=> 1–2+3-a+b=5 and 1-(-2)+3+a+b=19=> -a+b=5–1+2–3 and 1+2+3+a+b=19=> -a+b=3 and a+b=19–1–2–3=> -a+b=3 and a+b=13Adding these two equations,we get-a+b+a+b=3+13=> 2b=16=> 2b/2=16/2=> b=8Putting b=8 in a+b=13 , we geta+8=13=> a=13–8=> a=5Therefore, a=5 and b=8 .
How can I find the number of integers that satisfy the inequality [math]x^2 -6x +3 < 0[/math]?
Well you will first need to factorise the function. By applying formula method, we get the roots as[math]x = \dfrac{6 ± \sqrt(24)}{2}[/math][math]x = 3+\sqrt6 ; 3-\sqrt6[/math][math](x - 3 - \sqrt6) . (x - 3 + \sqrt6) < 0[/math]Now we will simply use an implication which we obtain from the wavy curve method which states that the quadratic will have a negative value for all values lying between the two roots (where coefficient of x² is positive)[This can be understood in a form that if you observe the graph of a quadratic equation, we know it’s an upward parabola if coeffcient of x² is positive. That means that if the roots are the point where the graph cuts the X-Axis then that implies that we will obtain the vertex/minima below x axis. And since it is below x axis, that means value of y will be negative.(something like this)Another way is that one factor has to be positive and the other has to be negative. So that implies that x has to be lesser than the solution which has a greater value but greater than the solution which has a lessee value]So this equation will be satisfied for all x lying between the roots 3+√6 and 3-√6.Now we dont know the exact value of √6 but we know that it will lie between 2 and 3. Well since (2.5)² = 6.25 and (2.4)² = 5.76 so then √6 will lie between 2.4 and 2.5.So 3+√6 > 5 and 3-√6 < 1.So the integer values will be 1, 2, 3, 4, 5.So that gives a solution of 5 integers.Hope it helps.
Solving Polynomial Equations Algebra II (High school level)?
1. Determine the zeros of f(x) = x3 – 3x2 – 16x + 48. (1 point) 2. Determine the zeros of f(x) = x3 – 12x2 + 28x – 9. (1 point) 3. Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18. (1 point) 4. Find the polynomial function with roots 1, –2, and 5. (1 point) 5. Find the polynomial function with roots 1, 7, and –3 of multiplicity 2. (1 point) 6. Find the polynomial function with roots 11 and 2i. (1 point) 7. Using complete sentences, describe how you would verify the zeros of a function. Use your description to verify one of the problems submitted for the assessment. (2 points) 8. Using complete sentences, describe your strengths and weaknesses in solving equations. (2 points)
What is a numerical coefficient?
Aswini's example is accurate, but his definition is not quite precise. A numerical coefficient is defined as a fixed number that is multiplied to a variable. So in the expression3x + 2y + 4as Aswini says, 3 is the coefficient of x, 2 is the coefficient of y, but, in my example, 4 is not a coefficient, as it is not being multiplied by a variable. It is a fixed value.Coefficients come up commonly when converting units of measures, as in these approximate conversions between metric and standard:1 KG = 2.2 LB meansX(kg) = 2.2*Y(lb)2.54cm = 1 inchX(cm) = 2.54*Y(inches)A more complex example is the conversion from Fahrenheit to Centigrade. One needs both a coefficient to adjust for the difference in the size of the unit of measure, and a fixed value to adjust for the zero pointX degrees centigrade = (5/9)*(Y Degrees Fahrenheit - 32)The fraction 5/9 is the coefficient, and 32 is the fixed value. (The fixed value must be adjusted before the coefficient is applied to match the zero points before adjusting for the size of the unit of measure.)Pi (approx. 3.14159) is a coefficient that shows up in a lot of mathematics and physics.In graphing, a coefficient of a variable (X) is related to the slope of the line. The fixed value determines the value of Y when X is zero. The term coefficient can apply to variables raised to different values. The generic quadratic equation is: Ax^2 + Bx + C = 0, where A and B are coefficients.
What are the roots of the equation [math] x^4 - 4x^3 + 8x^2 - 8x + 4 =0 [/math] ?
[math]x^4–4x^3+4x^2+4(x^2–2x+1)=0[/math][math](x(x-2))^2 +4(x-1)^2=0[/math][math](x^2–2x)^2+ 4(x-1)^2=0[/math][math][(x-1)^2–1 ]^2+4(x-1)^2=0[/math][math](x-1)^4–2(x-1)^2+1+4(x-1)^2=0[/math][math](x-1)^4+2(x-1)^2+1=0[/math][math][(x-1)^2+1 ]^2=0[/math]The roots are 1+i,1+i,1-i,1-i
How do I find the coefficient of [math]x^3[/math] in [math](1 - 2x + 3x^2 - 4x^3) ^{0.5}[/math]?
Markandeya Patowary's approach is very nice. You can also do this by brute force in software using the Python sympy library:>>> from sympy import *>>> x=symbols('x')>>> series(sqrt(1-2*x+3*x**2-4*x**3))1 - x + x**2 - x**3 - 3*x**4/2 - x**5/2 + O(x**6)