Answer:
Required Polynomial x²-x-12
Step-by-step Explanation:
Given that zeroes of a quadratic polynomial are -3 and 4
Let α = -3 and β= 4
Therefore, substituting the value α = -3 and β= 4 inf(x) = x²-(α +β)x+αβ, we get
f(x) = x²-( -3 + 4)x+(-3)(4)
= x²- x-12
Thus, x²-x-12 is the quadratic polynomial whose zeroes are -3 and 4.
Step-by-step explanation:
Let [tex] \sf \alpha \: and \: \beta [/tex]be the zeroes of polynomial .
[tex] \bf \: \alpha = - 3 \\ \\ \bf \beta = 4[/tex]
Now calculating sum of zeroes of the polynomial
[tex] \sf \: Sum \: of \: zeroes \: = \alpha + \beta \\ \\ \sf \: Sum \: of \: zeroes \: = - 3 + 4 \\ \\ \sf \: Sum \: of \: zeroes \: = 1[/tex]
again calculating the Product of zeroes of polynomial .
[tex] \sf \: Product \: of \: zeroes \: = \alpha \times \beta \\ \\ \sf \: Product \: of \: zeroes \: = - 3 \times 4 \\ \\ \sf \: Product \: of \: zeroes \: = - 12[/tex]
Now let's solve it further .
Let f (x) be the required ploynomial
As we know that ,
[tex] \underline{\boxed{\bf \: f(x) = {x}^{2} - (sum \: of \: zeroes)x \: + (product \: of \: zeroes)}}[/tex]
On substituting the value we get
[tex] \longrightarrow \sf \: f(x) \: = {x}^{2} - ( 1)x + ( - 12) \\ \\ \longrightarrow \sf \: f(x) \: = {x}^{2} -x - 12[/tex]
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Answers & Comments
Answer:
Required Polynomial x²-x-12
Step-by-step Explanation:
Given that zeroes of a quadratic polynomial are -3 and 4
Let α = -3 and β= 4
Therefore, substituting the value α = -3 and β= 4 inf(x) = x²-(α +β)x+αβ, we get
f(x) = x²-( -3 + 4)x+(-3)(4)
= x²- x-12
Thus, x²-x-12 is the quadratic polynomial whose zeroes are -3 and 4.
Verified answer
Step-by-step explanation:
Let [tex] \sf \alpha \: and \: \beta [/tex]be the zeroes of polynomial .
[tex] \bf \: \alpha = - 3 \\ \\ \bf \beta = 4[/tex]
Now calculating sum of zeroes of the polynomial
[tex] \sf \: Sum \: of \: zeroes \: = \alpha + \beta \\ \\ \sf \: Sum \: of \: zeroes \: = - 3 + 4 \\ \\ \sf \: Sum \: of \: zeroes \: = 1[/tex]
again calculating the Product of zeroes of polynomial .
[tex] \sf \: Product \: of \: zeroes \: = \alpha \times \beta \\ \\ \sf \: Product \: of \: zeroes \: = - 3 \times 4 \\ \\ \sf \: Product \: of \: zeroes \: = - 12[/tex]
Now let's solve it further .
Let f (x) be the required ploynomial
As we know that ,
[tex] \underline{\boxed{\bf \: f(x) = {x}^{2} - (sum \: of \: zeroes)x \: + (product \: of \: zeroes)}}[/tex]
On substituting the value we get
[tex] \longrightarrow \sf \: f(x) \: = {x}^{2} - ( 1)x + ( - 12) \\ \\ \longrightarrow \sf \: f(x) \: = {x}^{2} -x - 12[/tex]