Please help me factorise 5x^2 -14x +8?
(5x-4)(x-2)
Factor the trinomial 5x^2 + 14x - 3?
This is very similar to the one you asked 10mins ago
Find the value of a and b so that x+1 and x-2 are the factors of x^4+ ax^3-3x^2+2x+b?
X=-1 and x=2X^4+ax^3–3x^2+2x+b=0Put x=-1(-1)^4+a(-1)^3–3(-1)^2+2(-1)+b=01-a-3–2+b=0-a+b-4 =0-a+b=4=>(1)Putx=216+8a-12+4+b=08a+b=-8=>(2)equ.(1)-(2)-9a=12a =-4/3Put equ.1 in a= -4/3-(-4/3)+b=44/3+b=4b=8/3
Using the greatest common factor...?
Greatest Common Factor, or GCF, is something interesting that is used in future math problems to come. If you know your multiplication tables: 7, 14, 21, 28, 35... are units of 7. Let's take a look at your first problem: 7x + 14y - 28 As you can see, each one is a unit of 7. You can take out the 7 from each through division, then wrap them in parenthesis. Finally, multiply the parenthesis by 7. 7x + 14y - 28 (x + 2y - 4) * 7 Do you see something there? Let's look at our multiplication table: 2, 4, 6, 8, 10... There's another GCF! We can take that out in the same method. (x + 2y - 4) * 7 (x + 2(y - 2)) * 7 There you go! Question 2: y^2 - 10y + 21 Let's follow the multiplications table for variables. X, 2X, 3X, 4X... X*X (x^2)... x^3... x^4 So, we know that: y * y = y^2 Then we can divide two of the bits by Y. y^2 - 10y + 21 y(y - 10) + 21 Alternatively, you may be in the class that tells you to go the extra mile. This one is easy, though. (y + A)(y + B) This is your format. If you simplify it, you'll get the following: y^2 + (A + B)y + AB Which is pretty simple, right? So, find what each bit is. y^2 - 10y + 21 -10 = (A + B) 21 = AB So, -3 * -7 = 21, and -7 - 3 = -10, that means: A = -3 B = -7 (Y - A)(Y - B) (Y - 3)(Y - 7) Question 3: Difference of squares is the easiest of them all to use. Take your problem: x^2 - 49 Find the square root of both x 7 Now put them in our previous format: (X + A)(X + B) (X + 7)(X + 7) Change the either positive sign to a negative sign. (X + 7)(X - 7) And, you're done! Question 4: Factor completely using the tools of the previous 3 questions. 2x^3 - 12x^2 - 14x 2x(x^2 - 6x - 7) 2x((x - 7)(x + 1)) To check any of the answers, distribute, simplify, and see if it matches the starting equation.
Factor: y^2 - 49 + 14x - x^2?
Starting with: y² - 49 + 14x - x² Group the x terms and the constant together: y² + {-49 + 14x -x²} Factor out a -1: y² - {x² - 14x + 49} Notice that the x grouping is a perfect square of the form x² - 2ax + a² y² - {x - 7}² Notice that the above is of the form a² - b² where a = y and b = x - 7) so this factors into (a - b)(a + b): (y - {x - 7})(y + {x - 7}) Distributing the minus sign through the {}s and dropping them in the second factor: (y - x + 7})(y + x - 7)
How do I find the value of a and b with the polynomial [math]f(x)=2x^3+ax^2-bx+3[/math] which has a factor [math]x+3[/math], and when divided by [math]x-2[/math], has a remainder of [math]15?[/math]
Let 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-18 + 3a + b +1 = 0, or3a + b = 17 … (1).When divided by (x-2) the polynomial yields a remainder of 15. So deduct 15 from the polynomial to get 2x^3 +ax^2 - bx + 3 - 15, or2x^3 +ax^2 - bx -12. This polynomial should be divisible by (x-2) without leaving a remainder. Hence x=2. Substitute 2 for x in2x^3 +ax^2 - bx -12, to get16 + 4a - 2b - 12 = 0, or4a - 2b = -4, or2a - b = -2 … (2).Add (1) and (2) to get5a = 15 or a = 3.Put that value of 3 for a in (1) to getb = 17 - 3a, or17 -9 = 8.Hence a = -3 and b = 8.Check: Re-write f(x) as 2x^3 +3x^2 - 8x + 3.If x = -3, f(3) = -54 +27 +24 + 3 = 0.If x = 2, f(2) = 16 +12 -16 + 3 = 15 which is the remainder as mentioned above.Hence a = -3 and b = 8.