Math, asked by SuryanshGupta1267, 4 months ago

Factorise ax^4 - 27ax/y^3

Answers

Answered by Anonymous

Factoring a3 - b3

An expression of the form a3 - b3 is called a difference of cubes. The factored form of a3 - b3 is (a - b)(a2 + ab + b2):

(a - b)(a2 + ab + b2) = a3 - a2b + a2b - ab2 + ab2 - b3 = a3 - b3

For example, the factored form of 27x3 - 8 (a = 3x, b = 2) is (3x - 2)(9x2 + 6x + 4).

Similarly, the factored form of 125x3 -27y3 (a = 5x, b = 3y) is (5x - 3y)(25x2 +15xy + 9y2).

To factor a difference of cubes, find a and b and plug them into (a - b)(a2 + ab + b2).

Factoring a3 + b3

An expression of the form a3 + b3 is called a sum of cubes. The factored form of a3 + b3 is (a + b)(a2 - ab + b2):

(a + b)(a2 - ab + b2) = a3 + a2b - a2b - ab2 + ab2 + b3 = a3 - b3.

For example, the factored form of 64x3 + 125 (a = 4x, b = 5) is (4x + 5)(16x2 - 20x + 25).

Similarly, the factored form of 343x3 + y3 (a = 7x, b = y) is (7x + y)(49x2 -7xy + y2).

To factor a sum of cubes, find a and b and plug them into (a + b)(a2 - ab + b2).

You can remember these two factored forms by remembering that the sign in the binomial is always the same as the sign in the original expression, the first sign in the trinomial is the opposite of the sign in the original expression, and the second sign in the trinomial is always a plus sign.

Factoring ax3 + bx2 + cx + d

ax3 + bx2 + cx + d can be easily factored if = First, group the terms: (ax3 + bx2) + (cx + d ). Next, factor x2 out of the first group of terms: x2(ax + b) + (cx + d ). Factor the constants out of both groups. This should leave an expression of the form d1x2(ex + f )+ d2(ex + f ). We can add these two terms by adding their "coefficients": (d1x2 + d2)(ex + f ).

Previous sectionProblemsNext pageFactoring Polynomials of Degree 3 page 2

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