Factorise (cubic Polynomial)
6x^3-5x^2-17x+6
Answers
Solution :-
checking x = 1,2, _____ we get,
At x = 1 :-
→ 6x^3-5x^2-17x+6
→ 6*(0)³ - 5*(0)² - 17*0 + 6
→ 0 - 0 - 0 + 6
→ 6 = Not Possible .
At x = 2 :-
→ 6x^3-5x^2-17x+6
→ 6*(2)³ - 5*(2)² - 17*2 + 6
→ 6*8 - 5*4 - 34 + 6
→ 48 - 20 - 28
→ 48 - 48
→ 0.
Hence, we can say That, (x - 2) will be Factor of Given Polynomial . . (As Remainder is 0).
So, Dividing The Polynomial by (x - 2) now, we get,
x - 2 ) 6x³ - 5x² - 17x + 6 ( 6x² + 7x - 3
6x³ - 12x²
7x² - 17x
7x² - 14x
-3x + 6
-3x + 6
0
Factorising The Quotient Now :-
→ 6x² + 7x - 3
→ 6x² + 9x - 2x - 3
→ 3x(2x + 3) - 1(2x + 3)
→ (2x + 3)(3x - 1)
Hence, Factors of cubic Polynomial 6x^3-5x^2-17x+6 is (x-2)(2x+3)(3x - 1).
Question-
Factorise (cubic Polynomial)
6x^3-5x^2-17x+6
________________________
Solution-
The remainder theorem tells us that if f(c)=r for some polynomial f(x), then r is the remainder when f(x) is divided by x−c.
In our case, c is 2 for the first part, and −2 for the second.
We can now substitute these values into f(x), giving
f(+2)f(−2)=28,=0.
Therefore the remainder when f(x) is divided by x−2 is 28, and the remainder on dividing by x+2 is 0.
(iii) factorise f(x) completely.
Since the remainder of f(x) divided by x+2 is zero, we know x+2 is a factor of f(x).
Check the factor theorem if you need to…
We now know that f(x) is of the form (x+2)(ax2+bx+c), where a, b and c are constants to be found. This means
f(x)=6x3+5x2−17x−6=(x+2)(ax2+bx+c).
We can find a, b and c by considering the coefficients of the different powers of x.
The only cubic term on the right hand side is ax3, so we can see by comparing it with the left hand side that a=6.
The constant term is 2c, which must equal −6, and so c=−3.
There’ll be two square terms in the expansion of the right hand side, namely 12x2 and bx2. Therefore
12+b⟹b=5=−7.
We can check this by looking at the linear terms; on the right hand side, we get 2bx and −3x, so
2b−3⟹2b⟹b=−17=−14=−7,as above.
Therefore we can write f(x)=(x+2)(6x2−7x−3).
The quadratic bracket here can be factorised by inspection, or by using the quadratic formula if we’re stuck. We find that
f(x)=(x+2)(3x+1)(2x−3).
Or we could use polynomial division…
Although the question doesn’t ask for this, click below for a plot of y=f(x), with the three roots indicated.
Show Plot
Given that the expression ax3−x2+bx+18 is exactly divisible by x2+x−6, find the value of a and b.
Let’s factorise; x2+x−6=(x+3)(x−2).
So we know that (x−2) and (x+3) are factors of the above cubic expression, and so if we substitute in x=2 or x=−3 to this cubic, the result has to be zero.
This leads to the following pair of equations:
8a+2b+14−27a−3b+9=0⟹4a+b+7=0=0⟹−9a−b+3=0.(1)(2)
We can solve these by adding them together:
−5a+10⟹a=0=2,
and substituting into (1), we find
4×2+b+7⟹b=0=−15.
We can check this by substituting both a and b into (2):
−9×2−(−15)+3=0.
We therefore find a=2, b=15, and that the full cubic function is 2x3−x2−15x+18.