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# Algebra/i just dont get these questionsi need help

Question
Find the remainder when t + 1 is divided into

14.)  P(t) = t5 + t4 + t3 + t2 + t + 1

Find the remainder when z + 2 is divided into

15.)  P(z) = z5 + 2z4 + z3 + 2z2 + z + 2

Solve using the given root:

16.)  2x3 + 9x2 + 7x - 6 = 0; -2

Solve using the given root:

17.)2z3 + z2 - 8z + 3 = 0; 1.5

Find a polynomial equation with integral coefficients that has the roots:

18.)-2, 2, -3

Find a polynomial equation with integral coefficients that has the roots:

19.)-2, -i, i

Find a polynomial equation with integral coefficients that has the roots:

20.)1, -1.5, 2

Solve given the two roots:

21.)2x4 - 5x3 - 11x2 + 20x + 12; -2, 3

Find an equation with integral coefficients that has the given roots:
22.)1, -3, -4

The way to do these is like long division, as sin 372 divided by 6 is the same as
3*10^2 + 7*10^1 + 2 divided by 6.  Looking at 6 into 3 gives nothing, so we add the next term.
Since the next term is 6, we have 37 divided by 6, and that is 6, remainder 1.
Adding the last term gives 12 divided by 6 is  2.
Overall, the answer of 372/6 = 62.

Note that this says we always look at the division of the highest power.

It look like a number following the letter is an exponent,
as in x4 means x to the 4th.  It is usually seen as x^4, but I'll go with x4.
Using this ...

14.)  P(t) = t5 + t4 + t3 + t2 + t + 1; divide by t+1
The highest power is 1, since we have 1t5.
If we take (t+1)t4, we get t5 + t4.  That leaves us with t3 + t2 + t + 1.

This highest power is 1, since we have 1t3.
If we take (t+1)t3, we get t3 + t2.  That leaves us with t + 1
.
Since we are dividing by t+1, the answer to that is 1 and the remainder is 0.
A summary of what we did gives (t+1)(t4 + t2 + 1).

15.)  P(z) = z5 + 2z4 + z3 + 2z2 + z + 2; divide by z+2.
The first number is z4, for z4(z+2) = z5 + 2z4.
Subtracting that from P(z) cancels the 1st two terms, so we have z3 + 2z2 + z + 2 left.

The highest power is z3, and z3/z is z2.  Taking z2(z+2) gives z3 + 2z2.
Subtracting that from z3 + 2z2 + z + 2 leaves us with z + 2.

Dividing z+2 by z+2 gives 1, so putting this together gives (z + 2)(z4 + z2 + 1).

Solve using the given root:
16.)  2x3 + 9x2 + 7x - 6 = 0; -2
Since we are given x=-2, that means x+2 = 0, and we can divide by x+2.

Since 2x3 / x = 2x2, and 2x2(x+2) = 2x3 + 4x2, subtracting this off gives
{since 2-2=0 and 9-4=5} 5x2 + 7x - 6.

Since 5x2/x = 5x, and 5x(x+2) = 5x2 + 10, subtracting this off gives
{ since 5-5=0 and 7-10=-3 } -3x - 6.

Dividing the by x+2 gives -3.

This means we get (x+2)(2x2 + 5x - 3).

17.)2z3 + z2 - 8z + 3 = 0 with a root of 1.5 means we can divide by z - 1.5.
Since the first term is 2z3, this means multiply z - 1.5 by 2z2, giving 2z3 - 3z2.
Subtracting 2z3 - 3z2 from 2z3 + z2 - 8z + 3 gives 2z3 + z2 - 8z + 3 - (2z3 - 3z2).

Note that minus a minus is a plus, so we have + 3z2 at the end.
This gives 4z2 - 8z + 3.  Dividing 4z2 by z gives 4z, so 4z(z-1.5) = 4z2 - 6z,
We have 4z2 - 8z + 3 - 4z2 + 6z = -2z + 3.

Note that z - 1.5, when multiplied by -2, gives -2z + 3.
Subtracting this off gives 0, so the overall is (z - 1.5)(2z^2 + 4z - 2).

18.) To find a polynomial with the roots -2, 2, -3 means multiply (x+2)(x-2)(x+3).
Note that (x-2)(x+2) = x2 - 4, and (x2 - 4)(x+3) gives x3 + 3x2 - 4x - 12.

19.) To have roots -2, -i, i means multiply (x+2)(x+i)(x-i).
Note that i = squareroot(-1).
Multiplying the last two terms, (x+i)(x-i), gives x2 + 1.
Multiplying (x+2)(x2 + 1) gives x3 + 2x2 + x + 2.

20.) To have roots 1, -1.5, 2, we need (x-1)(x+1.5)(x-2).
Taking the first and the last gives (x-1)(x-2) = x2 - x - 2x + 2 = x2 - 3x + 2.
Multiplying (x2 - 3x + 2)(x - 2) gives x3 - 3x2 + 2x - 2x2 + 6x - 4 = x3 - 5x2 + 8x - 4.

21.) Given that 2x4 - 5x3 - 11x2 + 20x + 12 has the two roots -2 and 3
means we can divide by (x+2) and (x-3).

Dividing by (x+2) gives (x+2)(2x3 - 9x2 + 7x + 6).
Dividing 2x3 - 9x2 + 7x + 6 by x-3 gives (x-3)(2x2 - 3x - 2).

The total answer is then (x+2)(x-3)(2x2 - 3x - 2).

22.) To have roots 1, -3, and -4, multiply (x-1)(x+3)(x+4).
Since (x-1)(x+3) = x2 + 2x - 3, and (x2 + 2x - 3)(x+4) = x3 + 6x2 + 5x - 12,
that's the answer.  This can be checked by putting in 1, -3, and -4 for x.
Note that to do this easier, put these values in ((x + 6)x + 5)x - 12.
For example, for x = -3, we know -3 + 6 is 3.
Taking this times - 3 gives -9.
Taking -9 + 5 gives -4.
Taking -4 times -3 gives 12.
Taking 12 - 12 is 0.

Algebra

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#### Scott A Wilson

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Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.

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I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.

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