# Synthetic Division

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When you are dividing by a linear factor, you don't "have" to use long polynomial division; instead, you can use synthetic-division, which is much quicker.

**Example :**Find the remainder of x

^{2}+ 5x + 6 and x-1.

Write the coefficients ONLY inside an upside-down division symbol: | |

Make sure you leave room inside, underneath the row of coefficients, to write another row of numbers later. Put the test zero, x -1=0 ⇒x = 1, at the left: | |

Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the division symbol: | |

Multiply this carry-down value by the test zero, and carry the result up into the next column: | |

Add down the column: | |

Multiply the previous carry-down value by the test zero, and carry the new result up into the last column: | |

Add down the column: This last carry-down value is the remainder. |

**So remainder is 12.**

3) x

^{3}+7x

^{2}+7x -15

**Solution :**

**As sum of all the coefficient ( 1 + 7 + 7 -15 )= 0 so (x -1 ) is one of the factor**

After getting one factor find the other factors by

**Synthetic - Division.**

As at the bottom of the line, there are 3 numbers so degree of the polynomial will be (3 -1 =2) and the numbers are 1,8,15.So the polynomial will be

x

^{2}+ 8x + 15

Now find the factors of 15 in such a way that the addition and subtraction of that factors will be 8x.

So the

**factors of 15 = 5 and 3**

x

^{3}+7x

^{2}+7x -15 = (x-1)(x + 5)(x + 3) are the factors.

**Polynomial**

• Degree of the Polynomial

• Zeros of Polynomial

• Remainder Theorem

• Find remainder by Synthetic Division

• Rational root test in Polynomial

• Solved Examples on Polynomial

• Degree of the Polynomial

• Zeros of Polynomial

• Remainder Theorem

• Find remainder by Synthetic Division

• Rational root test in Polynomial

• Solved Examples on Polynomial

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