{"id":244253,"date":"2025-07-04T22:41:49","date_gmt":"2025-07-04T22:41:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=244253"},"modified":"2025-07-04T22:41:51","modified_gmt":"2025-07-04T22:41:51","slug":"use-synthetic-division-to-find-the-quotient-and-remainder-if-the-first-polynomial-is-divided-by-the-second","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/use-synthetic-division-to-find-the-quotient-and-remainder-if-the-first-polynomial-is-divided-by-the-second\/","title":{"rendered":"\u00a0Use synthetic division to find the quotient and remainder if the first polynomial is divided by the second"},"content":{"rendered":"\n
 Use synthetic division to find the quotient and remainder if the first polynomial is divided by the second<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Of course. Here is the correct answer and a detailed explanation for the math problem.<\/p>\n\n\n\n
quotient<\/strong> \u2014 12x\u00b3 + 6x\u00b2 – 8x – 4
remainder<\/strong> \u2014 0<\/p>\n\n\n\n
Explanation<\/h3>\n\n\n\n
The problem asks to use synthetic division to divide the polynomial 12x\u2074 – 11x\u00b2 + 2 by x – 1\/2. Synthetic division is a simplified method of polynomial division, specifically used when the divisor is a linear expression of the form x – c.<\/p>\n\n\n\n
Step 1: Set up the Division<\/strong><\/p>\n\n\n\n
First, identify the coefficients of the dividend polynomial, 12x\u2074 – 11x\u00b2 + 2. It’s crucial to list the coefficients for every power of x in descending order, from the highest power down to the constant term. If a term is missing, we must use a zero as a placeholder. In this case, the x\u00b3 and x terms are missing.<\/p>\n\n\n\n
The polynomial can be rewritten as: 12x\u2074 + 0x\u00b3 – 11x\u00b2 + 0x + 2.
The coefficients are: 12, 0, -11, 0, 2.<\/p>\n\n\n\n
Next, identify the value of c from the divisor, x – c. For the divisor x – 1\/2, the value of c is 1\/2.<\/p>\n\n\n\n
We set up the synthetic division problem by writing c to the left and the coefficients of the dividend to the right:Generated code<\/p>\n\n\n\n
1\/2 | 12   0   -11   0   2\n    |\n    -----------------------<\/code><\/pre>\n\n\n\n
Step 2: Perform the Synthetic Division<\/strong><\/p>\n\n\n\n
    \n
  1. Bring down the first coefficient (12) below the line.<\/li>\n\n\n\n
  2. Multiply this number (12) by\u00a0c\u00a0(1\/2):\u00a012 * (1\/2) = 6. Write this result under the second coefficient (0).<\/li>\n\n\n\n
  3. Add the numbers in the second column:\u00a00 + 6 = 6.<\/li>\n\n\n\n
  4. Repeat the process: multiply the new number (6) by\u00a0c:\u00a06 * (1\/2) = 3. Write this under the third coefficient (-11) and add:\u00a0-11 + 3 = -8.<\/li>\n\n\n\n
  5. Continue this pattern:\u00a0-8 * (1\/2) = -4. Write this under the next coefficient (0) and add:\u00a00 + (-4) = -4.<\/li>\n\n\n\n
  6. For the final step:\u00a0-4 * (1\/2) = -2. Write this under the last coefficient (2) and add:\u00a02 + (-2) = 0.<\/li>\n<\/ol>\n\n\n\n
    The completed division looks like this:Generated code<\/p>\n\n\n\n
    1\/2 | 12   0   -11    0    2\n    |      6      3   -4   -2\n    --------------------------\n      12   6     -8   -4    0<\/code><\/pre>\n\n\n\n
    Step 3: Interpret the Result<\/strong><\/p>\n\n\n\n
    The numbers in the bottom row represent the coefficients of the quotient and the remainder. The last number, 0, is the remainder. The other numbers, 12, 6, -8, -4, are the coefficients of the quotient polynomial.<\/p>\n\n\n\n
    The degree of the quotient is always one less than the degree of the dividend. Since the dividend was a 4th-degree polynomial, the quotient will be a 3rd-degree polynomial. Using the coefficients from our result, the quotient is 12x\u00b3 + 6x\u00b2 – 8x – 4.<\/p>\n\n\n\n
    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
     Use synthetic division to find the quotient and remainder if the first polynomial is divided by the second The Correct Answer and Explanation is: Of course. Here is the correct answer and a detailed explanation for the math problem. quotient \u2014 12x\u00b3 + 6x\u00b2 – 8x – 4remainder \u2014 0 Explanation The problem asks to use synthetic division to […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-244253","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244253","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=244253"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244253\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}