Factoring Polynomial Expressions: A Comprehensive Approach

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Factoring Polynomial Expressions: A Comprehensive ApproachFactor the following expression completely. If the polynomial is prime, then state this as youranswer.49x square root–84xy + 36y square root = (7x–6y)²Answer:Let's go step bystep to factor the given expression:Expression:49x2−84xy+36y249x^2-84xy + 36y^2Step 1: Recognize the structureThe given expression is a quadratic trinomial. We can try to factor it by recognizing it as aperfect square trinomial.A perfectsquare trinomial takes the form:a2−2ab+b2=(a−b)2a^2-2ab + b^2 = (a-b)^2Where:•a2a^2 is the first term,•2ab2ab is the middle term,•b2b^2 is the last term.Step 2: Compare terms with the perfect square trinomial formWe can try to rewrite49x2−84xy+36y249x^2-84xy + 36y^2 to see if it fits the perfectsquare trinomial form:•The first term is 49x249x^2, which is the square of 7x7x (since (7x)2=49x2(7x)^2 =49x^2).•The last term is 36y236y^2, which is the square of 6y6y (since (6y)2=36y2(6y)^2 =36y^2).•The middle term is −84xy-84xy, which is −2×7x×6y-2\times 7x\times 6y, or −84xy-84xy.So, the expression does indeed match the form of a perfect square trinomial.Step 3: Factor the perfect square trinomialSince we have:49x2−84xy+36y2=(7x−6y)249x^2-84xy + 36y^2 = (7x-6y)^2Final Answer:The factored form of 49x2−84xy+36y249x^2-84xy + 36y^2 is:(7x−6y)2(7x-6y)^2Factor the following expression completely. If the polynomial is prime, then state this as youranswer.64x tothe third–1 = (4x-1)(16x² +4x +1)Answer:Let's factor the given expression:Expression:64x3−164x^3-1Step 1: Recognize the patternThis is adifference of cubes, since it fits the formula:a3−b3=(a−b)(a2+ab+b2)a^3-b^3 = (a-b)(a^2 + ab + b^2)In this case:•a=4xa = 4x (since (4x)3=64x3(4x)^3 = 64x^3)

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•b=1b = 1 (since 13=11^3 = 1)Step 2: Apply the difference of cubes formulaSubstitute a=4xa = 4x and b=1b = 1 into the formula:64x3−1=(4x−1)((4x)2+(4x)(1)+12)64x^3-1 = (4x-1)\left((4x)^2 + (4x)(1) + 1^2\right)Step 3: Simplify the terms inside the parenthesesNow, let's simplify each term inside the second factor:(4x)2=16x2(4x)^2 = 16x^2 (4x)(1)=4x(4x)(1) = 4x 12=11^2 = 1So the factored expression becomes:64x3−1=(4x−1)(16x2+4x+1)64x^3-1 = (4x-1)(16x^2 + 4x + 1)Final Answer:The completely factored form of 64x3−164x^3-1 is:(4x−1)(16x2+4x+1)(4x-1)(16x^2 + 4x + 1)Since the second factor, 16x2+4x+116x^2 + 4x + 1, cannot be factored further (it is aquadratic with no real roots), this is the final factored form.Factor the following expression completely, if the polynomial is prime, then state this as youranswer.36y square root + 60y + 25 = (6x+5)²Answer:Let's examine and factor the expression:Expression:36y2+60y+2536y^2 + 60y + 25Step 1: Recognize the patternThis expression is aperfect square trinomial. A perfect square trinomial follows the form:a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a + b)^2We need to identify aa and bb such that:a2=36y2andb2=25a^2 = 36y^2\quad\text{and}\quad b^2 = 25Step 2: Find aa and bb•For a2=36y2a^2 = 36y^2, we take the square root of 36y², which gives a=6ya = 6y.•For b2=25b^2 = 25, we take the square root of 25, which gives b=5b = 5.Step 3: Check the middle termThe middle term of a perfect square trinomial is 2ab2ab. Let’s check if it matches the middleterm in our expression:2ab=2(6y)(5)=60y2ab = 2(6y)(5) = 60yThis matches the middle term in the original expression.Step 4: Write the factored formSince it is a perfect square trinomial, the expression can be factored as:36y2+60y+25=(6y+5)236y^2 + 60y + 25 = (6y + 5)^2Final Answer:The completely factored form is:(6y+5)2(6y + 5)^2Use factoring techniques to solve the following quadratic equation.z square root + z–2 = 0z = 1 or z =-2

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Answer:Let's solve the quadratic equation step by step:Given equation:z2+z−2=0z^2 + z-2 = 0Step 1: Factor thequadratic expressionWe need to factor the quadratic expression z2+z−2z^2 + z-2.We are looking for two numbers that multiply to −2-2 (the constant term) and add up to 11(the coefficient of zz).The two numbers are 22 and −1-1, because:•2×−1=−22\times-1 =-2•2+(−1)=12 + (-1) = 1So we can factor the quadratic as:z2+z−2=(z+2)(z−1)z^2 + z-2 = (z + 2)(z-1)Step 2: Set each factor equal to 0Now that we have factored the equation, we set each factor equal to 0:(z+2)=0or(z−1)=0(z + 2) = 0\quad\text{or}\quad (z-1) = 0Step 3: Solve for zz1.From z+2=0z + 2 = 0, we get: z=−2z =-22.From z−1=0z-1 = 0, we get: z=1z = 1Final Answer:The solutions to the equation are:z=1orz=−2z = 1\quad\text{or}\quad z =-2Factor the following expression completely, if the polynomial is prime, then state this as youranswer.27x to the third + 8y to the third = (3x + 2y)(9x²-6xy + 4y²)Answer:Let's factor the given expression step by step.Givenexpression:27x3+8y327x^3 + 8y^3This is a sum of cubes, which can be factored using thesum of cubes formula:a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2-ab + b^2)Step 1: Identify aa and bbIn this case, we can rewrite 27x327x^3 and 8y38y^3 ascubes:•27x3=(3x)327x^3 = (3x)^3•8y3=(2y)38y^3 = (2y)^3So, a=3xa = 3x and b=2yb = 2y.Step 2: Apply the sum of cubes formulaUsing the sum of cubes formula:a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2-ab + b^2)Substituting a=3xa = 3x and b=2yb = 2y:27x3+8y3=(3x+2y)((3x)2−(3x)(2y)+(2y)2)27x^3 + 8y^3 = (3x + 2y)((3x)^2-(3x)(2y) +(2y)^2)Step 3: Simplify the terms inside the parenthesesNow, simplify each term in the second factor:•(3x)2=9x2(3x)^2 = 9x^2•(3x)(2y)=6xy(3x)(2y) = 6xy•(2y)2=4y2(2y)^2 = 4y^2So, the factorization becomes:27x3+8y3=(3x+2y)(9x2−6xy+4y2)27x^3 + 8y^3 = (3x + 2y)(9x^2-6xy + 4y^2)

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