how to solve cubic quadratic equations
Answers
Step-by-step explanation:
Now that we know how to factorise cubic polynomials, it is also easy to solve cubic equations of the form ax3+bx2+cx+d=0.
WORKED EXAMPLE 13: SOLVING CUBIC EQUATIONS
Solve: 6x3−5x2−17x+6=0
Find one factor using the factor theorem
Let f(x)=6x3−5x2−17x+6
Try
f(1)=6(1)3−5(1)2−17(1)+6=6−5−17+6=−10
Therefore (x−1) is not a factor.
Try
f(2)=6(2)3−5(2)2−17(2)+6=48−20−34+6=0
Therefore (x−2) is a factor.
Factorise by inspection
6x3−5x2−17x+6=(x−2)(6x2+7x−3)
Factorise fully
6x3−5x2−17x+6=(x−2)(2x+3)(3x−1)
Solve the equation
6x3−5x2−17x+6 =0 (x−2)(2x+3)(3x−1) =0 x=2 or x =
1
3
or x=−
3
2
Sometimes it is not possible to factorise a quadratic expression using inspection, in which case we use the quadratic formula to fully factorise and solve the cubic equation.
x=
−b±
√
b2−4ac
2a
WORKED EXAMPLE 14: SOLVING CUBIC EQUATIONS
Solve for x: 0=x3−2x2−6x+4
Use the factor theorem to determine a factor
Let f(x)=x3−2x2−6x+4
Try
f(1)=(1)3−2(1)2−6(1)+4=1−2−6+4=−3
Therefore (x−1) is not a factor.
Try
f(2)=(2)3−2(2)2−6(2)+4=8−8−12+4=−8
Therefore (x−2) is not a factor.
f(−2)=(−2)3−2(−2)2−6(−2)+4=−8−8+12+4=0
Therefore (x+2) is a factor.
Factorise by inspection
x3−2x2−6x+4=(x+2)(x2−4x+2)
x2−4x+2 cannot be factorised any further and we are left with
(x+2)(x2−4x+2)=0
Solve the equation
(x+2)(x2−4x+2) =0 (x+2)=0 or (x2−4x+2)=0
Apply the quadratic formula for the second bracket
Always write down the formula first and then substitute the values of a,b and c.
a=1;b=−4;c=2
x =
−b±
√
b2−4ac
2a
=
−(−4)±
√
(−4)2−4(1)(2)
2(1)
=
4±
√
8
2
=2±
√
2
Final solutions
x=−2 or x=2±
√
2