{"id":40254,"date":"2025-06-27T15:09:53","date_gmt":"2025-06-27T15:09:53","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40254"},"modified":"2025-06-27T15:09:54","modified_gmt":"2025-06-27T15:09:54","slug":"dividing-polynomials-find-the-quotient-and-the-remainder-by-using-long-division","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/dividing-polynomials-find-the-quotient-and-the-remainder-by-using-long-division\/","title":{"rendered":"Dividing Polynomials Find the quotient and the remainder by using long division."},"content":{"rendered":"\n

Dividing Polynomials Find the quotient and the remainder by using long division. Write your complete solution. 1. (x2 + 3x + 10) = (x + 2) 2. (10×3 + 5×2 + 75x – 40 ) = (2x + 1) 3. ( 3×4 \u2013 x3 + x – 2) = (3×2 + x + 1)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Sure! Below are the complete solutions with explanations for dividing polynomials using long division:<\/p>\n\n\n\n

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1. Divide (x\u00b2 + 3x + 10) by (x + 2)<\/strong><\/h3>\n\n\n\n

Step 1:<\/strong> Set up the division:x2+3x+10x+2\\frac{x^2 + 3x + 10}{x + 2}x+2×2+3x+10\u200b<\/p>\n\n\n\n

Step 2:<\/strong> Divide the first term:x2\u00f7x=xx^2 \u00f7 x = xx2\u00f7x=x<\/p>\n\n\n\n

Multiply the divisor by x:(x+2)(x)=x2+2x(x + 2)(x) = x^2 + 2x(x+2)(x)=x2+2x<\/p>\n\n\n\n

Subtract:(x2+3x+10)\u2212(x2+2x)=x+10(x^2 + 3x + 10) – (x^2 + 2x) = x + 10(x2+3x+10)\u2212(x2+2x)=x+10<\/p>\n\n\n\n

Step 3:<\/strong> Divide again:x\u00f7x=1x \u00f7 x = 1x\u00f7x=1<\/p>\n\n\n\n

Multiply:(x+2)(1)=x+2(x + 2)(1) = x + 2(x+2)(1)=x+2<\/p>\n\n\n\n

Subtract:(x+10)\u2212(x+2)=8(x + 10) – (x + 2) = 8(x+10)\u2212(x+2)=8<\/p>\n\n\n\n

Final Answer:<\/strong>Quotient=x+1,Remainder=8\\text{Quotient} = x + 1,\\quad \\text{Remainder} = 8Quotient=x+1,Remainder=8<\/p>\n\n\n\n

Thus:x2+3x+10=(x+2)(x+1)+8x^2 + 3x + 10 = (x + 2)(x + 1) + 8×2+3x+10=(x+2)(x+1)+8<\/p>\n\n\n\n

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2. Divide (10x\u00b3 + 5x\u00b2 + 75x – 40) by (2x + 1)<\/strong><\/h3>\n\n\n\n

Step 1:<\/strong> Setup:10×3+5×2+75x\u2212402x+1\\frac{10x^3 + 5x^2 + 75x – 40}{2x + 1}2x+110×3+5×2+75x\u221240\u200b<\/p>\n\n\n\n

Step 2:<\/strong> Divide:10×3\u00f72x=5x210x^3 \u00f7 2x = 5x^210×3\u00f72x=5×2<\/p>\n\n\n\n

Multiply:(2x+1)(5×2)=10×3+5×2(2x + 1)(5x^2) = 10x^3 + 5x^2(2x+1)(5×2)=10×3+5×2<\/p>\n\n\n\n

Subtract:(10×3+5×2+75x\u221240)\u2212(10×3+5×2)=75x\u221240(10x^3 + 5x^2 + 75x – 40) – (10x^3 + 5x^2) = 75x – 40(10×3+5×2+75x\u221240)\u2212(10×3+5×2)=75x\u221240<\/p>\n\n\n\n

Step 3:<\/strong> Divide:75x\u00f72x=37.575x \u00f7 2x = 37.575x\u00f72x=37.5<\/p>\n\n\n\n

Multiply:(2x+1)(37.5)=75x+37.5(2x + 1)(37.5) = 75x + 37.5(2x+1)(37.5)=75x+37.5<\/p>\n\n\n\n

Subtract:(75x\u221240)\u2212(75x+37.5)=\u221277.5(75x – 40) – (75x + 37.5) = -77.5(75x\u221240)\u2212(75x+37.5)=\u221277.5<\/p>\n\n\n\n

Final Answer:<\/strong>Quotient=5×2+37.5,Remainder=\u221277.5\\text{Quotient} = 5x^2 + 37.5,\\quad \\text{Remainder} = -77.5Quotient=5×2+37.5,Remainder=\u221277.5<\/p>\n\n\n\n

Thus:10×3+5×2+75x\u221240=(2x+1)(5×2+37.5)\u221277.510x^3 + 5x^2 + 75x – 40 = (2x + 1)(5x^2 + 37.5) – 77.510×3+5×2+75x\u221240=(2x+1)(5×2+37.5)\u221277.5<\/p>\n\n\n\n

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3. Divide (3x\u2074 \u2013 x\u00b3 + x \u2013 2) by (3x\u00b2 + x + 1)<\/strong><\/h3>\n\n\n\n

Step 1:<\/strong> Setup:3×4\u2212x3+x\u221223×2+x+1\\frac{3x^4 – x^3 + x – 2}{3x^2 + x + 1}3×2+x+13×4\u2212x3+x\u22122\u200b<\/p>\n\n\n\n

Step 2:<\/strong> Divide:3×4\u00f73×2=x23x^4 \u00f7 3x^2 = x^23×4\u00f73×2=x2<\/p>\n\n\n\n

Multiply:(3×2+x+1)(x2)=3×4+x3+x2(3x^2 + x + 1)(x^2) = 3x^4 + x^3 + x^2(3×2+x+1)(x2)=3×4+x3+x2<\/p>\n\n\n\n

Subtract:(3×4\u2212x3+x\u22122)\u2212(3×4+x3+x2)=\u22122×3\u2212x2+x\u22122(3x^4 – x^3 + x – 2) – (3x^4 + x^3 + x^2) = -2x^3 – x^2 + x – 2(3×4\u2212x3+x\u22122)\u2212(3×4+x3+x2)=\u22122×3\u2212x2+x\u22122<\/p>\n\n\n\n

Step 3:<\/strong> Divide:\u22122×3\u00f73×2=\u221223x-2x^3 \u00f7 3x^2 = -\\frac{2}{3}x\u22122×3\u00f73×2=\u221232\u200bx<\/p>\n\n\n\n

Multiply:(3×2+x+1)(\u221223x)=\u22122×3\u221223×2\u221223x(3x^2 + x + 1)\\left(-\\frac{2}{3}x\\right) = -2x^3 – \\frac{2}{3}x^2 – \\frac{2}{3}x(3×2+x+1)(\u221232\u200bx)=\u22122×3\u221232\u200bx2\u221232\u200bx<\/p>\n\n\n\n

Subtract:(\u22122×3\u2212x2+x\u22122)\u2212(\u22122×3\u221223×2\u221223x)=\u221213×2+53x\u22122(-2x^3 – x^2 + x – 2) – (-2x^3 – \\frac{2}{3}x^2 – \\frac{2}{3}x) = -\\frac{1}{3}x^2 + \\frac{5}{3}x – 2(\u22122×3\u2212x2+x\u22122)\u2212(\u22122×3\u221232\u200bx2\u221232\u200bx)=\u221231\u200bx2+35\u200bx\u22122<\/p>\n\n\n\n

Step 4:<\/strong> Divide:\u221213×2\u00f73×2=\u221219-\\frac{1}{3}x^2 \u00f7 3x^2 = -\\frac{1}{9}\u221231\u200bx2\u00f73×2=\u221291\u200b<\/p>\n\n\n\n

Multiply:(3×2+x+1)(\u221219)=\u221213×2\u221219x\u221219(3x^2 + x + 1)\\left(-\\frac{1}{9}\\right) = -\\frac{1}{3}x^2 – \\frac{1}{9}x – \\frac{1}{9}(3×2+x+1)(\u221291\u200b)=\u221231\u200bx2\u221291\u200bx\u221291\u200b<\/p>\n\n\n\n

Subtract:(\u221213×2+53x\u22122)\u2212(\u221213×2\u221219x\u221219)=169x\u2212179\\left(-\\frac{1}{3}x^2 + \\frac{5}{3}x – 2\\right) – \\left(-\\frac{1}{3}x^2 – \\frac{1}{9}x – \\frac{1}{9}\\right) = \\frac{16}{9}x – \\frac{17}{9}(\u221231\u200bx2+35\u200bx\u22122)\u2212(\u221231\u200bx2\u221291\u200bx\u221291\u200b)=916\u200bx\u2212917\u200b<\/p>\n\n\n\n

Final Answer:<\/strong>Quotient=x2\u221223x\u221219,Remainder=169x\u2212179\\text{Quotient} = x^2 – \\frac{2}{3}x – \\frac{1}{9},\\quad \\text{Remainder} = \\frac{16}{9}x – \\frac{17}{9}Quotient=x2\u221232\u200bx\u221291\u200b,Remainder=916\u200bx\u2212917\u200b<\/p>\n\n\n\n

Thus:3×4\u2212x3+x\u22122=(3×2+x+1)(x2\u221223x\u221219)+(169x\u2212179)3x^4 – x^3 + x – 2 = (3x^2 + x + 1)\\left(x^2 – \\frac{2}{3}x – \\frac{1}{9}\\right) + \\left(\\frac{16}{9}x – \\frac{17}{9}\\right)3×4\u2212x3+x\u22122=(3×2+x+1)(x2\u221232\u200bx\u221291\u200b)+(916\u200bx\u2212917\u200b)<\/p>\n\n\n\n

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Explanation <\/strong><\/h3>\n\n\n\n

Polynomial long division follows the same principle as numerical long division. The goal is to simplify a polynomial expression by dividing it by a lower-degree polynomial, resulting in a quotient and possibly a remainder. First, identify the leading term of the dividend and divide it by the leading term of the divisor. This gives the first term of the quotient. Multiply the entire divisor by this term and subtract it from the dividend, simplifying the expression. Repeat the process with the new polynomial until the degree of the remainder is less than that of the divisor.<\/p>\n\n\n\n

For example, in the first question, dividing x2+3x+10x^2 + 3x + 10×2+3x+10 by x+2x + 2x+2, the quotient x+1x + 1x+1 and remainder 888 indicate that after factoring out x+2x + 2x+2, the polynomial simplifies with a small leftover constant.<\/p>\n\n\n\n

In the second example, with a cubic divided by a linear binomial 2x+12x + 12x+1, the same steps apply but more care is needed with coefficients, especially decimals. The quotient includes 5×2+37.55x^2 + 37.55×2+37.5 with a remainder of \u221277.5-77.5\u221277.5.<\/p>\n\n\n\n

The third example demonstrates polynomial division with more complex expressions and fractions, showing how division can still proceed systematically by managing fractional coefficients. Even when fractions appear, the same process ensures accuracy.<\/p>\n\n\n\n

Understanding this technique helps in simplifying expressions, solving polynomial equations, and analyzing algebraic functions, especially in calculus and advanced algebra topics.<\/p>\n\n\n\n

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Dividing Polynomials Find the quotient and the remainder by using long division. Write your complete solution. 1. (x2 + 3x + 10) = (x + 2) 2. (10×3 + 5×2 + 75x – 40 ) = (2x + 1) 3. ( 3×4 \u2013 x3 + x – 2) = (3×2 + x + 1) The […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-40254","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40254","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=40254"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40254\/revisions"}],"predecessor-version":[{"id":40256,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40254\/revisions\/40256"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40254"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40254"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40254"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}