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Original
2026-01-01
Modified
2026-02-28
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<p>Let’s see the step-by-step method for synthetic division using an example. Divide \((2x^3 + 5x^2 - 3x + 4) {\text { by }} (x - 1) \)</p>
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<p>Let’s see the step-by-step method for synthetic division using an example. Divide \((2x^3 + 5x^2 - 3x + 4) {\text { by }} (x - 1) \)</p>
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<p><strong>Step 1: Check the polynomial</strong></p>
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<p><strong>Step 1: Check the polynomial</strong></p>
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<p>Check whether the given polynomial is in<a></a><a>standard form</a>. All terms should be arranged in<a>descending order</a>of powers.</p>
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<p>Check whether the given polynomial is in<a></a><a>standard form</a>. All terms should be arranged in<a>descending order</a>of powers.</p>
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<p>The given polynomial is \(2x^3 + 5x^2 - 3x + 4\). Now write down only the coefficients \(2x^3 + 5x^2 - 3x + 4 \) becomes 2, 5, -3, 4.</p>
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<p>The given polynomial is \(2x^3 + 5x^2 - 3x + 4\). Now write down only the coefficients \(2x^3 + 5x^2 - 3x + 4 \) becomes 2, 5, -3, 4.</p>
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<p>And the given divisor is \((x - 1)\), we solve \(x - 1= 0\) and get \(x = 1\). This makes dividing easier.</p>
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<p>And the given divisor is \((x - 1)\), we solve \(x - 1= 0\) and get \(x = 1\). This makes dividing easier.</p>
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<p><strong>Step 2: Set the synthetic division box</strong></p>
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<p><strong>Step 2: Set the synthetic division box</strong></p>
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<p>We have to create a division box with the divisor and the<a>dividend</a>. Divisor: 1 Dividend: 2, 5, -3, 4</p>
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<p>We have to create a division box with the divisor and the<a>dividend</a>. Divisor: 1 Dividend: 2, 5, -3, 4</p>
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<p>\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ \end{array} \)</p>
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<p>\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ \end{array} \)</p>
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<p><strong>Step 3: Bring down the first number</strong></p>
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<p><strong>Step 3: Bring down the first number</strong></p>
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<p>Just bring down the first number </p>
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<p>Just bring down the first number </p>
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<p>\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ &\downarrow\\ & &2 \end{array} \)</p>
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<p>\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ &\downarrow\\ & &2 \end{array} \)</p>
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<p><strong>Step 4: Multiply and add </strong></p>
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<p><strong>Step 4: Multiply and add </strong></p>
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<p>Multiply the first number that we brought down by the divisor and write the answer under the next<a>number</a>.</p>
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<p>Multiply the first number that we brought down by the divisor and write the answer under the next<a>number</a>.</p>
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<p>\(1 × 2 = 2\)</p>
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<p>\(1 × 2 = 2\)</p>
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<p>Now we have to write the 2 below the next number that is 5, and add both numbers together.</p>
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<p>Now we have to write the 2 below the next number that is 5, and add both numbers together.</p>
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<p>\(5 + 2 = 7\)</p>
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<p>\(5 + 2 = 7\)</p>
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<p>Now do the same multiplication and addition with the number 7</p>
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<p>Now do the same multiplication and addition with the number 7</p>
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<p>\(1 × 7 = 7\\ -3 + 7 = 4\)</p>
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<p>\(1 × 7 = 7\\ -3 + 7 = 4\)</p>
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<p>And again, repeat the steps with 4</p>
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<p>And again, repeat the steps with 4</p>
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<p>\(1 × 4 = 4\\ 4 + 4 = 8\)</p>
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<p>\(1 × 4 = 4\\ 4 + 4 = 8\)</p>
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<p>We get the number 8, which is the<a>remainder</a>, and the first three numbers will be the<a>quotient</a>.</p>
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<p>We get the number 8, which is the<a>remainder</a>, and the first three numbers will be the<a>quotient</a>.</p>
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<p>\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ &\downarrow &2 & 7 & 4 \\ & &2 & 7 & 4 & 8 \\ \end{array} \)</p>
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<p>\( \begin{array}{r|rrrr} & & & & \\ \hline & & & & \\ 1 & 2 & 5 & -3 & 4 \\ &\downarrow &2 & 7 & 4 \\ & &2 & 7 & 4 & 8 \\ \end{array} \)</p>
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<p><strong>Step 5: Write the final answer</strong></p>
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<p><strong>Step 5: Write the final answer</strong></p>
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<p>The numbers at the bottom, except the last number, are quotient coefficients. We started the polynomial with a degree of 3, now we have to go one power lower.</p>
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<p>The numbers at the bottom, except the last number, are quotient coefficients. We started the polynomial with a degree of 3, now we have to go one power lower.</p>
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<p>Therefore, the quotient becomes</p>
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<p>Therefore, the quotient becomes</p>
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<p>\(2x^2 + 7x + 4\)</p>
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<p>\(2x^2 + 7x + 4\)</p>
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<p>The last number, 8, is the remainder.</p>
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<p>The last number, 8, is the remainder.</p>
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<p>The final answer should be in the format of:</p>
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<p>The final answer should be in the format of:</p>
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<p>\({p(x)\over q(x)} = {Q(x) + {R\over q(x)}} \) \({{2x^3+5x^2-3x+4\over x-1}} = (2x^2 + 7x + 4) + {{8\over (x - 1)}} \)</p>
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<p>\({p(x)\over q(x)} = {Q(x) + {R\over q(x)}} \) \({{2x^3+5x^2-3x+4\over x-1}} = (2x^2 + 7x + 4) + {{8\over (x - 1)}} \)</p>
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