12 multiple-choice questions, progressively harder.
Evaluate (2x2−x+3)+(x2+4x−1)(2x^2 - x + 3) + (x^2 + 4x - 1)(2x2−x+3)+(x2+4x−1) at x=2x = 2x=2 by simplifying first.
Solution
Correct answer: A
First combine like terms.
(2x2+x2)+(−x+4x)+(3−1)=3x2+3x+2(2x^2 + x^2) + (-x + 4x) + (3 - 1) = 3x^2 + 3x + 2(2x2+x2)+(−x+4x)+(3−1)=3x2+3x+2
Now substitute x=2x = 2x=2.
3(2)2+3(2)+2=12+6+2=203(2)^2 + 3(2) + 2 = 12 + 6 + 2 = 203(2)2+3(2)+2=12+6+2=20
What is the coefficient of x2x^2x2 in (4x3−2x2+x)−(x3+3x2−5)(4x^3 - 2x^2 + x) - (x^3 + 3x^2 - 5)(4x3−2x2+x)−(x3+3x2−5)?
Correct answer: D
Distribute the minus sign and look only at the x2x^2x2 terms.
−2x2−3x2=−5x2-2x^2 - 3x^2 = -5x^2−2x2−3x2=−5x2
The coefficient of x2x^2x2 is −5-5−5.
A rectangle has length 2x+52x + 52x+5 and width x−2x - 2x−2. By how much does the length exceed the width?
Correct answer: C
"Exceeds by" means length minus width. Distribute the minus sign.
(2x+5)−(x−2)=2x+5−x+2=x+7(2x + 5) - (x - 2) = 2x + 5 - x + 2 = x + 7(2x+5)−(x−2)=2x+5−x+2=x+7
Forgetting to flip the −2-2−2 gives x+3x + 3x+3, the common error.
The polynomials ax2+3x−1ax^2 + 3x - 1ax2+3x−1 and 2x2+bx+42x^2 + bx + 42x2+bx+4 add to 7x2−x+37x^2 - x + 37x2−x+3. What are aaa and bbb?
Match like terms. The x2x^2x2 terms give a+2=7a + 2 = 7a+2=7 and the xxx terms give 3+b=−13 + b = -13+b=−1.
a=5,b=−4a = 5, \qquad b = -4a=5,b=−4
If P(x)=x2+1P(x) = x^2 + 1P(x)=x2+1 and Q(x)=x2−xQ(x) = x^2 - xQ(x)=x2−x, what is P(x)+Q(x)P(x) + Q(x)P(x)+Q(x)?
Add the two polynomials term by term.
(x2+x2)−x+1=2x2−x+1(x^2 + x^2) - x + 1 = 2x^2 - x + 1(x2+x2)−x+1=2x2−x+1
Which polynomial must be added to 3x2−4x+13x^2 - 4x + 13x2−4x+1 to obtain x2x^2x2?
The needed polynomial is x2−(3x2−4x+1)x^2 - (3x^2 - 4x + 1)x2−(3x2−4x+1).
x2−3x2+4x−1=−2x2+4x−1x^2 - 3x^2 + 4x - 1 = -2x^2 + 4x - 1x2−3x2+4x−1=−2x2+4x−1
Check: (3x2−4x+1)+(−2x2+4x−1)=x2(3x^2 - 4x + 1) + (-2x^2 + 4x - 1) = x^2(3x2−4x+1)+(−2x2+4x−1)=x2.
Simplify: 2(x2+3)+(x2−3)2(x^2 + 3) + (x^2 - 3)2(x2+3)+(x2−3), reading 2(x2+3)2(x^2 + 3)2(x2+3) as (x2+3)+(x2+3)(x^2 + 3) + (x^2 + 3)(x2+3)+(x2+3).
Correct answer: B
Doubling x2+3x^2 + 3x2+3 is adding it to itself, giving 2x2+62x^2 + 62x2+6. Then add x2−3x^2 - 3x2−3.
2x2+6+x2−3=3x2+32x^2 + 6 + x^2 - 3 = 3x^2 + 32x2+6+x2−3=3x2+3
Simplify: (x2+2x+3)−(x2+2x+3)+(x2+2x+3)(x^2 + 2x + 3) - (x^2 + 2x + 3) + (x^2 + 2x + 3)(x2+2x+3)−(x2+2x+3)+(x2+2x+3).
The first two polynomials are identical and cancel, leaving the third.
(x2+2x+3)−(x2+2x+3)=0(x^2 + 2x + 3) - (x^2 + 2x + 3) = 0(x2+2x+3)−(x2+2x+3)=0
So the whole expression equals x2+2x+3x^2 + 2x + 3x2+2x+3.
What is the leading coefficient of (2x3+4x2)−(2x3−x2+x)(2x^3 + 4x^2) - (2x^3 - x^2 + x)(2x3+4x2)−(2x3−x2+x)?
Subtract first; the x3x^3x3 terms cancel, so the leading term changes.
2x3+4x2−2x3+x2−x=5x2−x2x^3 + 4x^2 - 2x^3 + x^2 - x = 5x^2 - x2x3+4x2−2x3+x2−x=5x2−x
The leading term is now 5x25x^25x2, so the leading coefficient is 555.
Simplify: (5x−3)−(2x+1)−(x−4)(5x - 3) - (2x + 1) - (x - 4)(5x−3)−(2x+1)−(x−4).
Subtract left to right, distributing each minus sign.
(5x−3)−(2x+1)=3x−4(5x - 3) - (2x + 1) = 3x - 4(5x−3)−(2x+1)=3x−4
Then subtract the last polynomial.
3x−4−x+4=2x3x - 4 - x + 4 = 2x3x−4−x+4=2x
For what value of kkk does (kx2+2x)+(3x2+x)(kx^2 + 2x) + (3x^2 + x)(kx2+2x)+(3x2+x) have degree 111?
The sum is (k+3)x2+(2x+x)=(k+3)x2+3x(k + 3)x^2 + (2x + x) = (k + 3)x^2 + 3x(k+3)x2+(2x+x)=(k+3)x2+3x. For degree 111 the x2x^2x2 term must vanish.
k+3=0 ⇒ k=−3k + 3 = 0 \;\Rightarrow\; k = -3k+3=0⇒k=−3
Then the sum is 3x3x3x, which has degree 111.
Let A=3x2+5A = 3x^2 + 5A=3x2+5 and B=x2−2x+5B = x^2 - 2x + 5B=x2−2x+5. What is the constant term of A−BA - BA−B?
Distribute the minus sign across BBB. Its +5+5+5 flips to −5-5−5 and cancels the +5+5+5 in AAA.
3x2+5−x2+2x−5=2x2+2x3x^2 + 5 - x^2 + 2x - 5 = 2x^2 + 2x3x2+5−x2+2x−5=2x2+2x
The constant term is 000; it cancelled.
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