Please help! Thanks in advance.

Find the values of a and b such that 1 and 4 are zeros of
[tex] f(x) = 2x^4-5x^3-14x^2+ax+b [/tex]

Respuesta :

mbh292
In this problem, we need to plug in the given x values for [tex] f(x)=0 [/tex] and find a and b.

When we plug in 1, we get: [tex]2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0[/tex]

Simplify:
[tex]2 - 5 - 14 + a + b = 0[/tex]
[tex] - 17 + a + b = 0[/tex]
[tex]a + b = 17[/tex]

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
[tex]2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0[/tex]
[tex]512 - 320 - 224 + 4a + b = 0[/tex]
[tex] - 32 + 4a + b = 0[/tex]
[tex]4a + b = 32[/tex]

Now we have our second one. We can combine them:
[tex]a + b = 17 \\ 4a + b = 32[/tex]

I use elimination method which is easier here.

Multiply the top equation by -1: [tex] - a - b = - 17 \\ 4a + b = 32[/tex]

Add them up:
[tex]3a = 15[/tex]

Simplify:
[tex]a = 5[/tex]

Now we have a, we can plug in one of those equations to find b:
[tex]5 + b = 17[/tex]
[tex]b = 17 - 5[/tex]
[tex]b = 12[/tex]

So, the answers are [tex]a=5[/tex] and [tex]b=12[/tex] .