Algebra II Help?

Question

Find all of the zeros of the following function.

p(x)=6x^4+22x^3+11x^2-38x-40


I'm having all sorts of problems figuring this one out. Can someone walk me through it step-by-step? Thanks.

Answer 1

You need to look at factors of 6 and 40 to find potential solutions. Any rational solutions will be a factor of 40 divided by a factor of 6 (positive or negative). Synthetic division is your best bet, but due to the difficulty of aligning it here I won't write it out.

Testing a few possibilities, we find x = -2 is a zero:
p(2) = 6(16) + 22(-8) + 11(4) - 38(-2) - 40 = 96 - 176 + 44 + 76 - 40 = 0. So divide through by (x+2) to get
p(x) = (x+2)(6x^3 + 10x^2 - 9x - 20)

Let q(x) = 6x^3 + 10x^2 - 9x - 20. Now we need factors of 20 divided by factors of 6.
We find x = 4/3 is a solution: q(4/3) = 6(64/27) + 10(16/9) - 9(4/3) - 20 = (384 + 480 - 324 - 540)/27 = 0. So we get
q(x) = (3x - 4) (2x^2 + 6x + 5)
The discriminant of this quadratic is 6^2 - 4(2)(5) = 36 - 40 = -4 < 0, so there are no additional zeros.

So the only zeros of p(x) are -2 and 4/3.

Answer 2

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Answer 3

try to group; suppose: (2x^2+ax+b) (3x^2+cx+d) = 6x^4 +(3a+2c)x^3 +(3b+2d+ac)x^2 +(cb+ad)x +bd; also suppose a, b, c, d are integer;
hence 3a+2c=22; 3b+2d+ac=11; cb+ad=-38; bd=-40;
♠ now 3a+2c=22, hence a must be even, b must be odd, c must be even, d must be even; and b or d must be negative;
♣ so 3(a/2)+c =11, then 3+8=11 and {a=2, c=8} or 9+2=11 and {a=6, c=2};
♦ now 3b+2d+ac=11, hence 3b +4(d/2) +5=0 or 3b +4(d/2) +1=0; thus 3b+1 must be divisible by 4; thus (3b+1)/4 +1 =-d/2 or (3b+1)/4 =-d/2, multiply by b,
♥ b(3b+1)/4 +b=20 or b(3b+1)/4 =20; b(3b+1)/4 +b=20 is non-consistent,
thus b=5 and d=-8; since(♦) 3b+2d+ac=11, then ac=12, so(♣) a=6 and c=2;
check: (2x^2+6x+5) (3x^2+2x-8) =6xxxx +(18+4)xxx +(15+12-16)xx +(10-48)x –40 OK!
☻2x^2+6x+5=0, no real roots!
☺3x^2+2x-8=0, x1=-2, x2=4/3;

Answer 4

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Answer 5

Gosh. I like math and this is a tough one...... Let me think on this a bit.