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(a) Show that $(x+3)$ is a factor of $3x^3 + 10x^2 + x^2 - 8x - 6$ - Scottish Highers Maths - Question 10 - 2019
Question 10
(a) Show that $(x+3)$ is a factor of $3x^3 + 10x^2 + x^2 - 8x - 6$. (b) Hence, or otherwise, factorise $3x^3 + 10x^2 + x^2 - 8x - 6$ fully.
Worked Solution & Example Answer:(a) Show that $(x+3)$ is a factor of $3x^3 + 10x^2 + x^2 - 8x - 6$ - Scottish Highers Maths - Question 10 - 2019
Step 1
Show that $(x+3)$ is a factor of $3x^3 + 10x^2 + x^2 - 8x - 6$.
Answer
To show that is a factor of the polynomial , we will use synthetic division or polynomial division. First, we will evaluate the polynomial at :
\ f(-3) &= 3(-3)^3 + 10(-3)^2 + (-3)^2 - 8(-3) - 6 \ &= 3(-27) + 10(9) + 9 + 24 - 6 \ &= -81 + 90 + 9 + 24 - 6 \ &= 36. \ ext{Since the result is } 0, ext{ we conclude that } (x + 3) ext{ is a factor.} \ ext{Thus, } (x + 3) ext{ is a factor of } 3x^3 + 10x^2 + x^2 - 8x - 6.\ ext{Therefore, we have shown that } (x + 3) ext{ is a factor.} ext{} \ ext{Alternatively, we can also perform polynomial long division to conclude the same.}Step 2
Hence, or otherwise, factorise $3x^3 + 10x^2 + x^2 - 8x - 6$ fully.
Answer
To factorise the polynomial , we can start by using the factor we found in part (a), which is , to perform polynomial division:
-
Divide:
- Performing synthetic or polynomial division of by , we find:
\2. Factor: - Next, we can factor the quadratic . - To find the roots of , we can use the quadratic formula:
x = rac{-b \pm ext{sqrt}(b^2 - 4ac)}{2a} where , , and .
= rac{-1 \\pm ext{sqrt}(1 + 24)}{6} = rac{-1 \\pm 5}{6}. This gives us:
- x = rac{4}{6} = rac{2}{3},
- x = rac{-6}{6} = -1.
Thus, the complete factorisation is:
The fully factorised form of the polynomial is: