Polynomials - A maths

bonkysleuth SG

Solve the equation 16y^3 + 4y^2 - 36y - 9 =0

I subbed in all the values ranging from -3 to 3 but none fits the equation so I can't really deduce the values of y. Is there some other solution?

Forbiddensinner SG

Solve the equation 16y^3 + 4y^2 - 36y - 9 =0

I subbed in all the values ranging from -3 to 3 but none fits the equation so I can't really deduce the values of y. Is there some other solution?

Hint : There are times you should try fraction values like -3/2, -1/2, 1/2 , 3/2

Opps, I just gave you two of the answers.

Lee012lee SG

Solve the equation 16y^3 + 4y^2 - 36y - 9 =0

I subbed in all the values ranging from -3 to 3 but none fits the equation so I can't really deduce the values of y. Is there some other solution?

Under the new syllabus of Add. Maths, students are allowed to use scientific calculators eg CASIO FX 95 SG PLUS with equation function to help them to find the first factor.

Steps

(1) Press Mode

(2) Choose 3 for equation

(3) Choose 4 for ax^3 + bx^2 + cx + d = 0

(4) Key in 16 for a

(5) Key in 4 for b

(6) Key in -36 for c

(7) key in -9 for d

The calculator will give the y answers to be 3/2, - 1/4, - 3/2.

Students should choose

y = 3/2

2y = 3

2y - 3 = 0

ie the first factor will be 2y - 3

Hence,

16y^3 + 4y^2 - 36y - 9 =0

(2y - 3)(8y^2 + 14y +3) = 0

(2y - 3)(4y + 1)(2y + 3) = 0

y = 3/2, - 1/4, - 3/2.

wee_ws SG

Hi,

Thanks for sharing a wonderful tip! I could relate it to H2 maths too :)

Cheers,
Wen Shih

bonkysleuth SG

Back to the topic again.

But my calculator is casio fx-82AU. I did consult my teacher regarding the use of calculator to find the factors but she said it isn't necessary. If that's the case, are they any other methods to solve such questions?

SBS n SMRT SG

den guess and check like mad....this is a short cut...want to use it or not up to you

TenSaru SG

Actually this question you can factorise 1st...

16y^3 + 4y^2 - 36y - 9 = 0

4y^2(4y+1) - 9(4y+1) = 0

(4y+1)(4y^2 - 9) = 0

(4y+1)(2y-3)(2y+3) = 0

Then solve

bonkysleuth SG

Actually this question you can factorise 1st...

16y^3 + 4y^2 - 36y - 9 = 0

4y^2(4y+1) - 9(4y+1) = 0

(4y+1)(4y^2 - 9) = 0

(4y+1)(2y-3)(2y+3) = 0

Then solve

That is an awesome method! Thanks!

Ok, there's another polynomial question which I'm pretty unclear. It involves having to sketch a graph.

The curve whose equation is y = (2x^2 + 3x -9)(x - k), where k is a constant has a turning point where x = -1.

(a)Calculate the value of k and the value of x at the other turning point on the curve

(b)Draw a rough sketch of the curve and find the set of values of x for which y > 0

For a , the answers are 9 and 6 respectively. However for drawing the graph, how do i know when to start/stop the limits along the x-axis? thank you.

Forbiddensinner SG

That is an awesome method! Thanks!

Ok, there's another polynomial question which I'm pretty unclear. It involves having to sketch a graph.

The curve whose equation is y = (2x^2 + 3x -9)(x - k), where k is a constant has a turning point where x = -1.

(a)Calculate the value of k and the value of x at the other turning point on the curve

(b)Draw a rough sketch of the curve and find the set of values of x for which y > 0

For a , the answers are 9 and 6 respectively. However for drawing the graph, how do i know when to start/stop the limits along the x-axis? thank you.

1) Find the turning points ( which you have done )

2) Find whether they are max. or min. point

3) Find intersections at y-axis and x-axis.

Presto, you have your graph.

And...what do you mean by the limits? I don't think there is any value of x undefined for this graph.

wee_ws SG

(b)Draw a rough sketch of the curve and find the set of values of x for which y > 0

However for drawing the graph, how do i know when to start/stop the limits along the x-axis? thank you.

Hi,

I believe you are referring to the range of values of x.

We just draw the graph in such a way that it shows the key features of the curve, i.e. all the turning points and x-intercepts.

Thanks!

Cheers,
Wen Shih