Sketch the polynomial function $y=x(x-1)^2(x-2)^3$,

 

Solve the inequality(LG Ex2-5B Q6f)

 

Quadratic inequalities using the graph (LG Ex2-5B Like Q2)

 

Like Q12 in LG Ex1-4B - remainder theorem and simultaneous equations

Question: When a polynomial $P(x)$ is divided by $(x+2)(x-4)$ the quotient is the polynomial $Q(x)$ and the remainder is $ax+b$.  Find $a,b$ if $P(-2)=3$ and $P(4)=2$.

Solution.

Using the factor theorem we get two equations for $a,b$ which we can solve to find $a,b$.

We have $$P(x)=(x+2)(x-4)Q(x)+ax+b$$

Therefore

 $P(-2)=3=-2a+b\implies -2a+b=3$  (1)

and

 $P(4)=2=4a+b\implies 4a+b=2$  (2)

Subtract (2)-(2) gives

$-1=6a\implies a=-\dfrac{1}{6}$

Then $b=3+2(-1/6)=2\dfrac{2}{3}=8/3$.

Conclude: $$a=-\dfrac{1}{6}\:\ \mbox{ and }\:\   b=\dfrac{8}{3}$$ .

Remember to check my working and tell me if I made an error!! (reward!)

Express $6x^3+35x^2+34x-40$ as a product of linear factors. (LG Ex1-4B Q7f)

 

Try factors of $40$ increasing order, usually a low value of $x$ will work.

Factors of $40$ are $1,2,4,5,,8,10,20,40$ and their negatives.

$P(-1)=-6+35-34-40\ne 0$

$P(1)=6+35+34-40\ne 0$

$P(-2)=-48+140-68-40\ne 0$

$P(-4)=-384+560-136-40=0\ $  !!  yay

Therefore by the factor theorem, $(x-(-4))=(x+4)$ is a factor of $P(x)$.

$\therefore P(x)=(x+4)(6x^2+bx-10)$.

I guessed the quadratics first and last terms as its obvious just looking at it!!

The $x$-term gives us $b$:  $4bx-10x=34x\implies b=11$

$\therefore P(x)=(x+4)(6x^2+11x-10)$.

Factorising the quadratic gives us

$\therefore P(x)=(x+4)(3x-2)(2x-5)$.