Simplify Calculator in the form you need

Yes. Simplifying, expanding and evaluating — with every step, the numerical check and practice problems — is free, with no sign-up. Each rule is named in the working, so you can see why a step is allowed, not just what the final answer is, and compare it with your own method.

It is the order of operations: Brackets, Orders (powers and roots), Division and Multiplication from left to right, then Addition and Subtraction from left to right. PEMDAS is the US name — Parentheses, Exponents, Multiplication/Division, Addition/Subtraction. Both give the same answer, for example $2+3\times (4-1{)}^2=29$2 + 3 × (4 - 1)^2 = 29.

Take out the largest square factor, so $\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$√(72) = √(36 × 2) = 6√(2), and add only like surds. To rationalise, multiply the top and bottom by the surd or by the conjugate: $\frac{2}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{2}$(2)/(√(5) - 1) × (√(5) + 1)/(√(5) + 1) = (√(5) + 1)/(2), because the bottom becomes $5-1=4$5 - 1 = 4. Exam questions usually want this exact form rather than a decimal.

Multiply every term in one bracket by every term in the other, then collect like terms. For $(2x+1)(x^2-3x+4)$(2x + 1)(x^2 - 3x + 4) that is six products, $2x^3-6x^2+8x+x^2-3x+4$2x^3 - 6x^2 + 8x + x^2 - 3x + 4, which collect to $2x^3-5x^2+5x+4$2x^3 - 5x^2 + 5x + 4. Choose Expand or type "expand" to see every product written out.

Because the original expression divides by something containing $x$x. For example $\frac{x}{x}=1$x/x = 1 only when $x\ne 0$x ≠ 0 — at $x=0$x = 0 the original is $\frac{0}{0}$0/0, which is undefined. A simplified answer is only equal to the original where the original exists, so those values are stated.

A negative index means "one over": $x^{-2}=\frac{1}{x^2}$x^(-2) = (1)/(x^2). A fractional index is a root: $a^{\frac{p}{q}}={\left(\sqrt[q]{a}\right)}^p$a^(p/q) = (root q of (a))^p, so $8^{\frac{2}{3}}=2^2=4$8^(2/3) = 2^2 = 4. Multiplying powers of the same base adds the indices, so $x^{\frac{1}{2}}\cdot x^{\frac{3}{2}}=x^2$x^(1/2) × x^(3/2) = x^2, and any non-zero base to the power $0$0 is $1$1.

Yes. Type a mixed number with a space, such as `2 1/3 + 1 1/2`: it is first written as an improper fraction ($\frac{7}{3}$7/3) and the answer is given as $\frac{23}{6}$23/6 and as the mixed number $3\frac{5}{6}$35/6. Factorials such as `5!` or `3! + 4!` are worked out in the "orders" stage of BODMAS.