Q&A for How to Solve Polynomials

The original equation was 5x + 2 = 0. Then 2 was subtracted from both sides of the equation in order to begin the process of solving for x. This resulted in 5x = -2.

You get x - 2 + 1/x.

Yes. To factor a trinomial, you must split it into a quadratic polynomial.

x^4 - x² = x²(x² - 1) = 0. Therefore, x² = 0, or x² - 1 = 0. If x² = 0, then x = 0. If x² - 1 = 0, then x = +/-1.

Add 2 to both sides of the equation.

It's not an equation, so it can't be "solved." In other words, no value for x can be found. However, if you just want to perform the multiplication, you'll get the product x^6 - x³ - 42.

Because x = 4, the remainder theorem states that P(4) = 0. So sub P(4) = 0, essentially then solve for A.

Because the equation has two unknown variables (y and j), it can't be solved. When you have two unknowns, you need two independent equations in those unknowns in order to solve for them.

Divide both sides of the equation by x: 25x² = 64. Then find the square root of both sides: +/- 5x = +/- 8. So x = +/- 8/5.

The sentence means, 6x^2 + x - 1 = 0. To factor a trinomial, figure out what two numbers for the coefficients multiply to get 6. 3 and 2 (as well as 1 and 6) work. Next, find out what two numbers for the constants multiply to equal -1. Only 1 and -1 work. Use FOIL to test the numbers, and you will see that (3x-1)(2x+1) is the factored form. Set 3x - 1 and 2x + 1 equal to zero. x can equal 1/3 or -1/2.

See Divide Polynomials .

You can't solve it unless it's an equation. If you were to set that expression equal to 0, then x² - 64 = 0. Add 64 to both sides: x² = 64, and x = +/- 8.

That's an equation in three unknowns (four if you count x as an unknown). To solve for three unknowns, you would need three independent equations. With just one equation in three unknowns, you could solve for just one unknown, and the solution would be in terms of the other two unknowns.

4^n = (2²)(4 ²) = (4)(16) = 64. By inspection, 4³ = 64, so n = 3.

(-6²)³ = (-6)^6 = +46,656.

First, factor out the a. What remains is the sum of two cubes: 8a + 125ax³ = a(8 + 125x³) = a[(2)³ + (5x)³]. Factoring the sum of two cubes yields this: a(2 + 5x)(2² - 10x + 25x²) = a(2 + 5x)(4 - 10x + 25x²).

That appears to be an equation in two unknowns (m and d). In order to solve such an equation, you would need two separate equations, each containing both unknowns. The only alternative is to solve for one of the unknowns in terms of the other. In this case, m = d - 2, or d = m + 2.

This equation is solved similarly to x² - 9x +18 = 0 = (x - 6)(x - 3). x^4 - 9x² + 18 = 0 = (x² - 6)(x² - 3).

Calculating (x3+ax2-7x+b)/(x-1) gives condition b=6-a. Calculating (x3+ax2-7x+b)/(x-2) gives condition b=6-4a. Solving both conditions give a=0 and b=6, we find polynomial x3-7x+6=(x-1)(x-2)(x+3)=0 results in solution x=-3,x=1 and x=2. Or, using property of roots in cubic equation: Since we know two roots of the polynomial and we know the minus times coefficient of x is equal to product of roots: r1*r2+r2*r3+r1*r3=-(-7) using r1=1 and r2=2 we have 1*2+2*r3+r3=-7 and we get r3=-3 coefficient of x2 is sum of roots r1+r2+r3=a this gives a= 1+2-3 so a=0 coefficient of constant is minus times product of roots -(r1*r2*r3)=b=-(1*2*-3) so b=6.

(a) Calculate (4x3-19x2+ax-14)/(x-2)=4x2-11x+a-22. We get a condition for a : -14=-2(a-22) and then solve for a=29. (b) f(x)=4x3-19x2+29x-14=(x-2)(4x2-11x+7), solving quadratic part of f(x): g(x)=4x2-11x+7 x=(4x-7)(x-1) and find x=7/4 and x=1 (also possible with a,b,c formula), solving for f(x) we find x=1, x=7/4 and x=2.

There is no standard form, because a polynomial can have any number of terms.

Use the answer that you got in the original equation to see if it still holds up. If you had the equation 3x + 2 = 5 and got x = 1, you can apply 3(1) + 2 = 5 to confirm your answer.

It's not possible to solve an equation expressed in two unknowns (x and y in this case).

4x³ + 3x = x(4x² + 3) = 0. Then x = 0, or (4x² + 3) = 0. In the latter case, 4x² = -3, x² = -¾, and x is the square root of a negative number, which is an "imaginary" number. If we restrict our answer to "real" numbers, x = 0.