MATHEMATICS: Synthetic Division, an Alternative Method?
Use synthetic division to find the quotient. ANSWERS only! No solutions needed. 1) (x^3 - 2x^2 + x + 18) / (x + 2) 2) (3x^2 - 2x - 4) / (x - 3) 3) (x^3 + 3x^2 -2x + 5) / (x + 2) 4) (x^3 + 4x - 7) / (x - 3) 5) (-x^4 + 7x^3 - 4x^2) / (x - 3) 6) (x^5 -4x^3 +5x^2 -5) / (x + 1) 7) (2x^3 + 5x^2 -4x -3) / (x+1) 8) (x^5 + 32) / (x + 2) 9) (a^4 – 16) / (a – 2) 10) (2a^4 + 5a^3 + 7a^2 -4a + 6) / (a + 3) Use synthetic division. ANSWERS only! No solutions needed. 11) (x^3 – 2x – 3x^2 – 5) / (x + 3) 12) (2a^3 + 3a^2 – 7) / (a + 1) 13) (y^3 + 3y^2 - 2y + 5) / (y + 1) 14) (2x^4 – x^3 - 18x^2 – 7) / (x + 3) 15) (3x^4 - 15x^3 + 2x - 10) / (x – 5) 16) (x^4 – 2x^2 – 7x + 6) / (x - 1)
Synthetic division (12x^4-56x^3+59x^2+9x-18) / (2x+1)?
I get a slightly different answer than you...the -3x^2 is -31x^2. Otherwise, the rest of the terms line up. This is what I did: Since 2x+1 is the same as 2(x+1/2) we can divide using synthetic division for the x+ 1/2 factor, and then divide everything by 2. *-1/2*___12__-56__59__9___-18 ____________-6___31__-45__18 _______12___-62__90__-36___0 The remainder is 0 We get a quotient of 12x^3-62x^2+90x-36. But we haven't divided by the 2 yet. 6x^3-31x^2+45x-18
Use long division to divide 3x^4-10x^3+2x^2-7x-1 by x^2-4x+3. What is the remainder?
You don’t need to perform long division in order to just find the remainder. We know that the remainder has the form [math]ax+b[/math] and that, for some polynomial [math]Q(x)[/math],[math]3x^4-10x^3+2x^2-7x-1=(x^2-4x+3)Q(x)+ax+b[/math]Since [math]x^2-4x+3=(x-1)(x-3)[/math], we can evaluate both sides at 1 and 3, getting[math]\begin{cases}-13=a+b \\ -31=3a+b\end{cases}[/math]and now it’s easy to find [math]a=-9[/math] and [math]b=-4[/math].
Without actual division, can you prove that ‘2x^4-6x^3+3x^2+3x-2’ is exactly divisible by ‘x^2-3x+2’?
If P(x)=2x^4–6x^3+3x^2+3x-2 is divisible byg(x)=x^2–3x+2=(x-2)(x-1) thenP(1)=0 and P(2)=0P(1)=2–6+3+3–2=8–8=)P(2)=2×16–6×8+3×4+3×2–2=32–48+12+6–2=50–50=0Therefore P(x) is divisible by g(x).
Solve for x: 18x to the 4th power+15x+15x to the 3rd power-34x squared-2=0?
Let f(x) = 18x^4 + 15x^3 - 34x^2 + 15x - 2 = 0 Graph it on a graphing calculator or computer. It looks like there are single zeros at x = -2 and x = 1/2 and a possible double zero near x = 0.3. Calculate f(-2) : = 18(-2)^4 + 15(-2)^3 - 34(-2)^2 + 15(-2) - 2 = 288 - 120 - 136 - 30 - 2 = 0 so, x = -2 is a zero, therefore, (x + 2) is a factor. Calculate f(1/2) : = 18(1/2)^4 + 15(1/2)^3 - 34(1/2)^2 + 15(1/2) - 2 = 18/16 + 15/8 - 34/4 + 7.5 - 2 = 0 so, x = 1/2 is a zero, therefore, (x - 1/2) is a factor. Now divide f(x) by (x + 2) using synthetic division. The answer is : 18x^3 - 21x^2 + 8x - 1 Now divide this by (x - 1/2). The answer is : 18x^2 - 12x + 2 = 2(9x^2 - 6x + 1) = 2(3x - 1)^2 Setting this equal to zero, shows the double zero is at x = 1/3. Thus, f(x) = 18x^4 + 15x^3 - 34x^2 + 15x - 2 = 2(x + 2)(x - 1/2)(3x - 1)^2 = (x + 2)(2x - 1)(3x - 1)^2 and the values of x when f(x) = 0 are -2, 1/2 and 1/3.
Solve the polynomial equation in order to obtain the first root....?
The rational root test yields possible roots of +/- 1, +/- 2, +/- 3, +/- 6, +/- 9, +/- 18. Use synthetic division with these possible values. It turns out that using synthetic division w/ -2 will work quite nicely, yielding the following factors: (x + 2)(x^2 - 9) This can then be easily factored to (x + 2)(x - 3)(x + 3). Note: You could have used factoring by grouping from the beginning and arrived at the same answer without use of synthetic division...
How do I solve Y=x^4+5x^3-4x^2+7x-2 using the first principles?
We cannot read your mind as to exactly what the question means…You give an equation in two variables, [math]x[/math] and [math]Y[/math]. (By the way, and why change from lower-case for the first variable to upper-case for the second variable??) Hence this equation will have infinitely many solutions, because for each real number [math]x[/math] the pair [math](x, Y) = (x, x^4 + 5 x^3 - 4 x^2 + 7x - 2)[/math] will be a solution.Did you perhaps mean, instead to solve the single-variable equation [math] x^4 + 5 x^3 - 4 x^2 + 7x - 2 = 0[/math] for [math]x[/math]? If so, I don’t think that has any simply expressed solutions. The “quartic formula” is applicable, of course, but would be very tedious to carry out by hand, I think.Here are the four solutions as computed by Mathematica:
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.