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Work through fractions, percentages, measurements, shapes, and useful calculations with examples.
Search this topicArrange both polynomials in descending powers, using zero placeholders for missing powers. Divide the leading term, multiply the whole divisor by that result, then subtract and repeat with the remainder. For (x² + 7x + 10)/(x + 2), start with x, subtract x² + 2x, then use +5. The quotient is x + 5 with no remainder.
Find the largest numerical factor and the lowest power of each variable shared by every term. In 15x³ + 25x², both terms contain 5x². Divide each term by it and write 5x²(3x + 5). Expand your brackets afterward to check both terms return. A variable missing from even one term cannot be part of the shared factor.
Group terms so each pair has a common factor, then look for a shared bracket. For xy + 4x + 3y + 12, write x(y + 4) + 3(y + 4). Both parts now contain y + 4, giving (x + 3)(y + 4). Expand to verify the result. If the brackets differ, inspect the grouping and signs before continuing.
For x² + bx + c, look for two numbers whose product is c and whose sum is b. For x² + 11x + 28, 4 and 7 work, so the factors are (x + 4)(x + 7). Expand them to check the middle term. If no integer pair works, do not force one; another factoring or equation-solving method may be needed.
The two numbers must have opposite signs because their product is negative. For x² + 2x − 35, the pair +7 and −5 multiplies to −35 and adds to +2. Therefore the factorization is (x + 7)(x − 5). The larger absolute value has the sign of the middle coefficient. Always check both the product and sum.
First put the equation equal to zero. If (x − 4)(x + 6) = 0, at least one factor must be zero, giving x = 4 or x = −6. Substitute both values into the original equation to check them. You cannot apply this rule directly when the product equals a nonzero number; rearrange and factor again first.
It silently assumes x is not zero. In x² = 8x, dividing by x gives x = 8 but hides the valid solution x = 0. Instead move everything to one side: x² − 8x = 0, then factor x(x − 8) = 0. Both zero and eight satisfy the original equation. Check before dividing by any variable expression.
Cancel common factors, not individual pieces joined by addition. For (x² − 16)/(x − 4), factor the numerator as (x − 4)(x + 4), then cancel the shared factor to get x + 4. The original still excludes x = 4. In contrast, you cannot simply cancel the x in (x + 4)/x because it is not a factor of the whole numerator.
Set the original denominator equal to zero and solve, then exclude those values. For (x + 1)/[(x − 3)(x + 5)], x cannot be 3 or −5. Record these restrictions before simplifying; a canceled factor does not restore an input that made the original denominator zero. A zero numerator is allowed whenever the denominator remains nonzero.
Collect variable terms on one side and constants on the other using the same operation on both sides. For 7x + 2 = 3x + 22, subtract 3x to obtain 4x + 2 = 22, then subtract 2 and divide by 4. The solution is x = 5. Substitution gives 37 on both sides, confirming it.
Distribute first, then combine like terms and isolate the variable. In 4(x − 2) + 3 = 19, expanding gives 4x − 8 + 3 = 19, so 4x − 5 = 19 and x = 6. Check in the unexpanded version: 4(6 − 2) + 3 = 19. Keep signs attached to terms when removing the brackets.
For 6x² + 13x + 6, multiply the first and last coefficients to get 36. Find two numbers multiplying to 36 and adding to 13: 9 and 4. Split the middle term to get 6x² + 9x + 4x + 6. Group this as 3x(2x + 3) + 2(2x + 3), giving (3x + 2)(2x + 3).
It leaves a smaller expression whose coefficients are easier to work with. In 12x² + 26x + 12, first take out 2 to get 2(6x² + 13x + 6). The inner trinomial factors as (3x + 2)(2x + 3), so keep the outside 2 in your final answer. Expanding all factors should recover the original polynomial.
Use a³ − b³ = (a − b)(a² + ab + b²). For x³ − 125, recognize 125 as 5³ and write (x − 5)(x² + 5x + 25). The signs in the second bracket are both positive. Multiplying the factors checks the cancellation of the middle powers; subtracting two cubes is different from subtracting two squares.
Use a³ + b³ = (a + b)(a² − ab + b²). Thus x³ + 216 = (x + 6)(x² − 6x + 36). The first bracket keeps the plus sign, while the middle term of the second bracket is negative. Check that both original terms really are cubes before using the formula, and expand to verify the result.
Factor numerators and denominators first, record excluded values, then cancel shared factors across the product. For [(x² − 9)/(x + 1)] × [(x + 1)/(x + 3)], factoring x² − 9 gives (x − 3)(x + 3). Cancel to obtain x − 3, still excluding x = −1 and −3 from the original denominators.
Multiply by the reciprocal of the divisor, then factor and simplify. For [x/(x + 2)] ÷ [(x − 1)/(x + 2)], the product becomes [x/(x + 2)] × [(x + 2)/(x − 1)] = x/(x − 1). Exclude x = −2, where the original fractions are undefined, and x = 1, where the divisor equals zero.
Write the relationship as y = kx, then calculate k = y/x using a known pair with x not zero. If y is 21 when x is 6, k is 3.5 and the relationship is y = 3.5x. Doubling x doubles y. A straight-line relationship with a nonzero intercept is not direct variation in this form.
Write y = k/x, so the product xy stays constant. If y = 8 when x = 9, k = 72. When x becomes 12, y = 72/12 = 6. Increasing x reduces y for a positive constant. Confirm the situation actually follows this model; many real-world quantities do not maintain a constant product.
The symbol √64 denotes the principal, nonnegative square root, which is 8. The equation x² = 64 asks for every number whose square is 64, so both 8 and −8 work. Write x = ±8 when solving that equation. Keep the square-root symbol’s meaning separate from the list of solutions to a squared-variable equation.
For real x, √(x²) equals |x|, the nonnegative magnitude of x. If x = −9, squaring gives 81 and taking the principal square root gives 9, not −9. You may replace √(x²) with x only when x is known to be nonnegative. Keeping the absolute-value bars prevents an unnoticed sign error when the variable can be negative.
Look for the largest perfect-square factor. For √180, write √(36 × 5), then take √36 outside to obtain 6√5. The remaining 5 has no perfect-square factor greater than 1. You can check numerically that both expressions have the same value. Finding a smaller square factor first also works, but may require another simplification step.
Separate the largest squared part and apply the principal-root rule. For √(25x⁶), the expression is √[(5x³)²], which equals |5x³| or 5|x³|. If the problem states x ≥ 0, that becomes 5x³. Without such a restriction, retain the absolute value so negative inputs still produce the required nonnegative square root.
Treat the shared radical like a common unit. For 4√7 + 9√7, add the coefficients to get 13√7. Do not add the numbers under the square-root signs. When the radicals differ, simplify each first; if their remaining radicands still differ, they cannot be combined into a single like term through ordinary addition.
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